Calling all math experts and Pizza Hut fans alike! National Pi Day is here and this is your chance to win free “pie,” that’s 3.14 years of Pizza Hut pizza (awarded in Pizza Hut® gift cards)! Take a look at the math problems below and provide your answer to Option A, B, or C in the comments section. Please be sure to note which you are trying to solve. Answers will be time stamped to determine the potential winner and participants can only win once.

Best of luck!

– Pizza Hut & John H. Conway

OPTION A: SOLVED – WINNERS WILL BE NOTIFIED WITHIN 24 HOURS

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

OPTION B: SOLVED – WINNERS WILL BE NOTIFIED WITHIN 24 HOURS

Our school’s puzzle-club meets in one of the schoolrooms every Friday after school.

Last Friday, one of the members said, “I’ve hidden a list of numbers in this envelope that add up to the number of this room.” A girl said, “That’s obviously not enough information to determine the number of the room. If you told us the number of numbers in the envelope and their product, would that be enough to work them all out?”

He (after scribbling for some time): “No.” She (after scribbling for some more time): “well, at least I’ve worked out their product.”

What is the number of the school room we meet in?”

OPTION C: YET TO BE SOLVED, No one has gotten this one exactly right yet! Hint: It helps to show your work!

My key-rings are metal circles of diameter about two inches. They are all linked together in a strange jumble, so that try as I might, I can’t tell any pair from any other pair.

However, I can tell some triple from other triples, even though I’ve never been able to distinguish left from right. What are the possible numbers of key-rings in this jumble?

NO purchase necessary to enter, win or claim a Prize.  Contest open only to eligible legal residents of the 48 contiguous U.S. and D.C. who are at least 18.  Void in Alaska, Hawaii, Puerto Rico and where prohibited.  Official Rules found at http://blog.pizzahut.com/wp-content/uploads/2016/03/Pi-Day-Rules-FINAL-03-03-16-2.pdf

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• March 14 at 7:01 am

3816547290 for problem A

• March 14 at 7:06 am

option c: 5

• March 14 at 7:54 am

Problem A: 3816547290

• March 14 at 8:04 am

• March 14 at 8:09 am

• March 14 at 8:12 am

Option a is 3816547290
Option b is 314
Option c is 3.1415926535897932……….

• March 14 at 8:18 am

Option A: 38165472

• March 14 at 9:40 am

Option A=3816547290
Option B=314
Option C=3.141592653589793238462643383279502884197169399375105820974944592307

• March 14 at 10:18 am

Question A: 3816547290

Question B: 314

Question C: 3.14

• March 14 at 10:30 am

• March 14 at 10:36 am

Option A: 3816547290

• March 14 at 11:04 am

3.141592653589793238462643383279502884197169399375105820974944592307…

• March 14 at 11:05 am

3816547290 for option A

• March 14 at 12:09 pm

A: 3816547290

• March 14 at 1:54 pm

Solution:
3 is divisible by 1
38 is divisible by 2
381 is divisible by 3
3816 is divisible by 4
38165 is divisible by 5
381654 is divisible by 6
3816547 is divisible by 7
38165472 is divisible by 8
381654729 is divisible by 9
3816547290 is divisible by 10
The even numbers descend 8,6,4,2
The odd numbers nearly ascend save for 3,1

• March 14 at 2:06 pm

3816547290

• March 14 at 2:26 pm

The answer to C is 3 or more.

• March 14 at 2:36 pm

Option A:
3816547290

• March 14 at 3:21 pm

Wrong! Obviously the answer must be an integer value.

• March 14 at 3:36 pm

• March 14 at 3:50 pm

Answer C is any multiple of 3 plus any multiple of 4
Option B is any room with a 1 or 0 in it

• March 14 at 3:55 pm

• March 14 at 4:26 pm

Answer To Question 3: Any multiple of six

• March 14 at 4:30 pm

Option C is 32

• March 14 at 5:40 pm

Problem C = 3

• March 14 at 6:07 pm

• March 14 at 6:11 pm

• March 14 at 6:17 pm

Answer to C: 6, 12, 18, 24, 30, 36…

• March 14 at 6:47 pm

The answer to problem C is 48.

• March 14 at 6:55 pm

Option C: 9

• March 14 at 6:56 pm

Problem C: The number of key rings can be any odd number greater than or equal to 3.

Proof: If you can tell the triples apart and you can’t tell your left from your right then the positions of the three keys must be the same left to right as they are right to left.

So if there are N key rings and the positions of the three keys in a triple are A, B, and C (A, B, and C each being a number 0 through N-1). If left to right the left most key is at position A then right to left that key moves to position N-1-A. For this key to still be in the same triple N-1-A must be either A, B, or C. Same for N-1-B and N-1-C.

If you reverse the key ring positions then the left most key will trade places with the left most and the middle one will stay in the middle. Mathematically, this can be written as

N-1-A = C
N-1-B = B
N-1-C = A

When solved this comes to

B = (N-1)/2 and C is dependent on the position of A.

This means that, for example, if there are 5 keys a triple you can tell apart from any other is first position (A=0), third position (B=2), and 5th position (C=4).

As for restrictions on N, since B must be a whole number (N-1)/2 must be a whole number which only happens when N is odd.

• March 14 at 7:17 pm

Option c: 13
Arrange the rings like the face of a clock. Place one ring with its center in the center of the clock. This is ring 13. Place a center of a ring at each of the 12 number positions on the clock face. These rings each link each other. The center ring links each of the 12 number positioned rings. With this arrangement every combination of 2 rings can form a line with a length of 2 inches between the centers of each ring. Thus no pair is distinguishable from another. However a set of 3 rings on the clock face are connected in a circular arrangement and although any set of three rings on the face are indistinguishable from each other any set of 3 rings that goes from a clock number through the center of the clock face and back to another clock number opposite to it forms a straight line and is distinguishable from the other mentioned set of 3 rings in the number positions. Since this arrangement of rings is circular on the perimeter and has a circular center, no left or right exists.
The diameter of the clock is 4 inches because we use a center ring of diameter 2 inches connected to the center position of 2 other rings (adding 1 inch for each of the other 2 rings) Since Circumference of a circle equals pi times diameter we have the clock circumference of 3.14 times 4 inches which is about 12.5inches. Since the radius of each ring is 1 inch we need 12 rings to complete the circumference of the clock face, spacing them on their centers. Therefore the total number of rings needed are 12 perimeter rings plus one center ring for a total of 13 rings.

• March 14 at 7:20 pm

Answer to C: Every odd multiple of 3 starting from 3.

• March 14 at 7:36 pm

There are 2 options to the total sum:
3 + 4a +6b where a is the number of paired rings divided by two and b is the number of triple rings divided by 2,
or 2 + 4a + 6b where a is again the number of paired rings divided by 2 and b is the number of triple rings divided by 2.

• March 14 at 7:38 pm

Option C: 13 rings.
Arrange the center of a ring at each of the 12 number positions of a circular clock face. Place one ring in the center of the circular clock. Each of the 12 perimeter rings will link with their adjacent neighbor and with the center ring. Any pair of 2 rings forms a line through their centers that is indistinguishable from any other 2 pairs of rings. Any 3 rings on the perimeter of the clock face will follow the arc of the circle and are indistinguishable from each other. 3 rings formed from the center ring and 2 opposite rings will form a line and is distinguishable from the set of three perimeter rings that follow the arc of the circle. The diameter of the clock circle is 4 inches. (2 inch diameter for the center circle and 1 inch radius length for each opposite perimeter ring) Since circumference = pi times diameter , the circumference of the clock face is 4inches times 3.14 which is about 12.5 inches. Since the radius of each ring on the perimeter of the circle is 1 inch, we need 12 rings on the perimeter of the clock face(positioned like the numbers on a clock) Therefore the total number of rings is 12 perimeter rings plus 1 center ring for a total of 13 rings. Also this shape has no identifiable left or right.

• March 14 at 7:42 pm

Option C: 114

• March 14 at 7:51 pm

• March 14 at 8:49 pm

Option B: 0

• March 14 at 9:05 pm

answer to c is 5 + 2x, or 5 and every odd number following it

• March 14 at 10:00 pm

• March 14 at 10:32 pm

5, 7, 9, 12, 14, 16, 19, 21, 23, etc.

• March 14 at 11:07 pm

Option C: The answer is 4. There are 3 rings connected to each 1 ring. Therefore, no matter how you look at the rings you can tell there are three rings.

• March 14 at 11:44 pm

C is equal to any triangular number divisible by 3 OR -3 plus any triangular number divided by 3. This can be expressed by the statements (1/2)n(n+1) and (1/2)n(n+1)-3 where n is an integer greater than or equal to 2 in the first case and greater than or equal to 3 in the second. This is under the assumption that “any other” and “another” are equivalent statements.

• March 14 at 9:04 am

• March 14 at 9:20 am

Option A: 3816547290

• March 14 at 10:19 am

3816547290 is the only one solution (I know I’m late though)

• March 14 at 10:49 am

• March 14 at 11:11 am

can’t find my post, so doing it again…

• March 14 at 11:18 am

A. 75891361915

• March 14 at 1:03 pm

A: 1472583690 works too!!

• March 14 at 2:00 pm

@Darren: your first 8 digits are not divisible by 8 because 836 is not divisible by 8. 836/8 = 104.5.

• March 14 at 5:07 pm

Òption C: at least 7

• March 14 at 9:43 am

Option A:
3816547290
Even though so many other people have predicted this number.

• March 14 at 12:48 pm

the answer to A is actually 3816547290.
the answer to B is actually 314.
the answer to C must be a range because it is asking for the least amount of keys to the most amount of keys so the answer must be 6 to 24.

• March 14 at 5:51 pm

answered again because i couldn’t find my original . c=3

• March 14 at 10:00 am

How are you reaching that number specifically?

• March 14 at 2:53 pm

I personally wrote a script that iterated over all permutations of 0-9. Not sure how everyone else did it. Sidenote: I’m very excited that the scripts name was “pizza.py”

#!/usr/bin/python

import itertools
digits = 10

def makeint(l, n):
number = 0
for a in range(n):
number = number + 10 ** (n-a-1) * l[a]
return number

def check(i):
number = makeint(i,digits)
for a in range(1,digits+1):
if makeint(i,a) % a != 0:
return False
return True

for i in itertools.permutations(range(digits)):
if i[0] != 0:
if check(i):
print makeint(i,digits)

========
\$ ./pizza.py
3816547290

• March 14 at 1:20 pm

A: 3816547290

• March 14 at 1:54 pm

option B is 314

• March 14 at 2:54 pm

B: 632
C:7

• March 14 at 3:22 pm

Problem A.) 3816547290

• March 14 at 4:43 pm

C is any integer ≥ 10 that is factorable to 2 and 3.
“pairs” is plural, meaning more than one pair. same as “triples”. Therefore, there must be at least 2 “pairs” and two “triples”, or (2*2)+(3*2), making 10. It has to be an integer, as there was no mention to any rings being cut to bits, and it has to be factorable by 2 and 3 to stick with the multiple rings.

• March 14 at 5:51 pm

Answer to a ia 3186547290 and answer to c is 24 there are two pair orr sets that triple from triple so each set has three below and those have three below them.

• March 14 at 5:58 pm

Option C is 9.

• March 14 at 6:25 pm

A: 3816547290
B.314
C: 3.141592

• March 14 at 6:26 pm

Option A is 3816547290 or 4998

• March 14 at 8:22 pm

Problem C: 19

• March 14 at 11:22 pm

1,130 circles don’t have a volume. 1 x 1 x 3.14 x 360

• March 14 at 10:20 am

Prove it and don’t give out answers, this ruins the point.

• March 14 at 11:07 am

Yes! These people are forgetting that math is just as much proofs as getting the answer. If you can’t show me a proof then why should I believe your answer.

• March 14 at 1:13 pm

Option A: 3816547290. Lots of fun Pizza Hut!!! I hope you do this again next year!!!! THANK YOU

• March 14 at 2:44 pm

I got that answer (to part A) by brute-forcing it. Pretty simple program, really: http://cpp.sh/74sna

• March 14 at 11:56 am

• March 14 at 1:33 pm

room # 1

• March 14 at 12:57 pm

A:3816547290
B:Room 314
C:3.14……….

• March 14 at 1:41 pm

C: = 6561 combinations

• March 14 at 3:47 pm

Option A: 3816547290

By the way. I am an admirer of Dr. Conway’s work!

• March 14 at 11:19 am

Option a: 1234567890

• March 14 at 6:49 pm

Option C is 10 plus a multiple of 3.
I see a peace symbol type thing and at the center of it a triple. The chain of rings attached to the center and the outer circle triples also. This makes the conditions of the problem met i think. or maybe some variation of this…

• March 14 at 11:29 am

How to comment

• March 14 at 11:32 am

Option a is 3816547290

• March 14 at 11:34 am

3816547290 is the answer to question A

• March 14 at 11:35 am

Question c is 3.141592

• March 14 at 12:39 pm

Option a: 3.14159
Option b: 27
Option c: 6

• March 14 at 1:16 pm

C: still working on it… difficult

• March 14 at 3:49 pm

I posted my A & B answers earlier. For C, I suspect you had
in mind that the number N of rings had to be any one of the following
(in the below, q denotes a prime power):

N=q+1 where q=1 mod 4.
N=q^3+1 with q odd.
N=(8^n-2^n)/2 with n>2.
N=(8^n+2^n)/2 with n>2.
N=176.
N=276.
N=q^(2*n) where (n,q) is not (1,2).

Do I have a proof? Well, not really.
Actually I’m quite unconvinced this is the right answer.
But it probably has something to do with it.
If you want to know more I can email my notes in which I allegedly derived my answers.

• March 14 at 2:10 pm

Option A : 3816547290

• March 14 at 5:16 pm

Problem a: 3816547290

• March 14 at 2:59 pm

For option A– 3816547290

• March 14 at 3:23 pm

Option a: 1,234,759,680

• March 14 at 11:29 pm

Option A: 3816547290

• March 14 at 3:50 pm

Option A: 3816547290

• March 14 at 4:04 pm

1010101010 problem a
No. Problem b
3 problem c

• March 14 at 4:14 pm

C. 23

• March 14 at 4:16 pm

C: 26 rings

• March 14 at 4:18 pm

8

• March 14 at 4:27 pm

Answer for Option C is 8.

• March 14 at 4:39 pm

C: 1

• March 14 at 4:58 pm

• March 14 at 5:13 pm

A:1

• March 14 at 5:41 pm

B
D
H
P

• March 14 at 5:43 pm

B
D
H
P

• March 14 at 6:10 pm

Problem C:15

• March 14 at 6:12 pm

Question C: 5 Rings

• March 14 at 6:19 pm

Problem C :15

• March 14 at 6:35 pm

Option c: any multiple of 6

Like 6, 12, 18, 24

• March 14 at 6:41 pm

Option c: any multiple of 6
6, 12, 18, 24, ect

• March 14 at 6:46 pm

• March 14 at 6:50 pm

Option C is any perfect square except 4. (9,16,25,36…)
They are arranged in the shape of a square

• March 14 at 7:01 pm

C: 2^k * 3k for k >=2

• March 14 at 7:26 pm

Option C: 48

• March 14 at 7:45 pm

Option C: 5
Firstly, the rings must be arranged in a circle like shape with n rings in the center. Since a pair of rings cant be discerned from each other, the diameter(in rings) must be greater than 2. But, a triple set of rings are distinguishable– although not left to right(that’s why it’s arranged in a circle). Thus, the diameter(in rings) must be 3. The rings on the outside of the ‘circle’ must not be connected because then the triple sets wouldn’t be distinguishable. So the rings must be set up in a + shape with 5 rings.

• March 14 at 7:53 pm

UPDATE:
C = 5 + 2n
n >/= 0 (n is greater or equal to 0)
Proof in post above except there can be any number of additional rings only connected to the center as long as the have a partener connect on the opposite side of them

• March 14 at 7:50 pm

69

• March 14 at 10:23 pm

Question C: An even number 6 or greater.

• March 14 at 7:58 pm

Option C: 440169

• March 14 at 8:03 pm

option C: any multiple of 5

• March 14 at 8:37 pm

7 and 22

• March 14 at 8:50 pm

The answer I just posted was for Problem C, 7 and 22

• March 14 at 10:35 pm

Option C is 10

• March 14 at 10:43 pm

Option C is 11

• March 14 at 8:51 pm

Option C: 9

• March 14 at 8:53 pm

C is 54.

• March 14 at 9:34 pm

option c: 4 rings
Place the center of three rings on the vertex of an equilateral triangle with side length 2. Link these rings with each other. Place a 4th ring with its center in the center of the triangle and link this to each vertex ring. Each ring will be linked to 3 other rings .All pairs will look the same. The three triples that go from a vertex to the center to another vertex will all look the same. However, the triple that connects the 3 vertex rings will look different from the three triples that are made of the center ring with 2 vertexes. This configuration has no left or right, has pairs that all look the same, and have one triple that is distinguishable from the other triples.

• March 14 at 10:07 pm

Option C: 28

• March 14 at 10:20 pm

Option C: Any number 3^n, n=1,2,3… etc (by 3 and 4 and beyond is a lot of keys logically, but still could be possible)

• March 14 at 10:47 pm

Option C: One possible number of rings is 9. If right and left are indistinguishable, a core circle shape makes direction irrelevant. If we have 3 center rings making a circle, the addition of 2 rings, which themselves are connected, upon each center one creates some distinguishable triplets; this also creates pairs that can’t be told apart from any other, as the connected rings could all be seen to be part of multiple pairs.

• March 14 at 10:51 pm

Option C: is 12

• March 14 at 11:41 pm

Option C Any number greater than 5 that can be achieved by using a center ring with each other ring attached just to it.

• March 14 at 11:54 pm

C is equal to any triangular number divisible by 3 OR -3 plus any triangular number divided by 3. This can be expressed by the statements (1/2)n(n+1) and (1/2)n(n+1)-3 where n is an integer greater than or equal to 2 in the first case and greater than or equal to 3 in the second. This is under the assumption that “any other” and “another” are equivalent statements.

• March 14 at 7:06 am

• March 14 at 7:49 am

I think that this website is a good way to know what pizza hut is all about you can learn more things about pizza hut foods and you can
eat different pizzas and the places .

• March 14 at 8:19 am

• March 14 at 10:26 am

Option A=3816547290
Option B=21
Option C= 5

• March 14 at 9:10 am

Option A: is 3816547290

• March 14 at 9:18 am

A: 3816547290

• March 14 at 10:54 am

B. 16

• March 14 at 11:54 am

• March 14 at 12:15 pm

Option C: palindromic numbers that are multiples of 6 –
can be paired – multiples of 2, can be tripled – multiples of 3, can’t tell left from right – palindromic. Examples: 66, 606, 636, 666, etc.

• March 14 at 7:09 am

optionC: 2

• March 14 at 7:28 am

C:9

• March 14 at 9:48 am

• March 14 at 7:12 am

This doesn’t work for n = 8

• March 14 at 7:39 am

It does work for n = 8: 38165472/8 = 4770684

• March 14 at 9:18 am

N must equal 2 or 5 since we are dealing with an integer with 10 digits. It must be divisible by n, for each n.

• March 14 at 12:32 pm

Actually, n is every integer from 1 to 10. The question is saying that, for each of those integers (1, 2, 3, 4, 5, 6, 7, 8, 9, 10), the first n digits of the number are divisible by n. So, the first 3 digits of the number are divisible by 3, the first 6 are divisible by 6, and so on:

(3 digits) 381 / 3 = 127
(6 digits) 381654 / 6 = 63609

• March 14 at 7:12 am

C. 3.14159265358979

• March 14 at 12:23 pm

Option a – 3816547290

• March 14 at 7:13 am

b. 2

• March 14 at 8:53 am

Option B is 4

• March 14 at 8:57 am

B. 4

• March 14 at 10:52 am

B: 100

• March 14 at 7:14 am

Option A = 3816547290

• March 14 at 1:25 pm

A: 3816547290

• March 14 at 7:14 am

Option C

• March 14 at 7:15 am

Option C:

Any multiple of 6.

I.e. 6, 12, 18, …..

• March 14 at 8:30 am

C
Any multiple of 6

• March 14 at 9:33 am

A is 3816547290

• March 14 at 10:52 am

Option B: is 6 that’s what I got

• March 14 at 7:17 am

problem A 1896547290

• March 14 at 8:29 am

Is not divisible by 7. 1896547/7= 270935.2857

• March 14 at 7:18 am

Room 12 – Option B

• March 14 at 8:33 am

Option B: Room #1.

• March 14 at 9:31 am

A . 3816547290
B . 314
C . 3.141592653589

• March 14 at 11:38 am

314

• March 14 at 2:18 pm

Option a: 3816547290

• March 14 at 11:40 am

3.141592653589

• March 14 at 11:42 am

3.141592653589

• March 14 at 3:17 pm

for option b the answer is 314 and option c is 3.14 and option a is 3.141592653

• March 14 at 12:08 pm

Question A: 3816547290
Question B: 314
Question C: 3.14

• March 14 at 1:47 pm

Option B: 314

• March 14 at 7:19 am

Cheater

• March 14 at 7:57 am

Get off my nutz

• March 14 at 9:09 am

out number be N. For all even n, N must divide an even number and hence ends in a multiple of 2. Since all the digits must be distinct, we can say that N must have an odd integer at all odd positions n. Since N divides 10, it must end in 0. Since the first 5 digits of N divides 5, and the 5th digit can no longer be 0, the 5th digit is 5. Now N is:
(odd) (even) (odd) (even) 5 (even) (odd) (even) (odd) 0
And we have 1, 2, 3, 4, 6, 7, 8, and 9 left over. Note that, for the first 4 digits to divide 4, since the 3rd digit is even, the 4th digit must be 2 or 6. Similarly, for the first 8 digits to divide 8, since the 7th digit is odd, the 8th digit must be 2 or 6. If the 8th digit is a 2, the 7th digit is a 3 or 7, and if the 8th digit is a 6, the seventh digit must be a 1 (note that a 5 could also work, but that we have already used the 5).
Now N looks like this:
(odd) 4/8 (odd) 2/6 5 4/8 (odd) 2/6 (odd)
Since 2 and 6 are used in the 4th and 8th index, we know that they cannot be used anywhere else. The next step is to try values.
If N starts with 1:
If N starts with 14:
If N starts with 14725:
N must start with 1472583. From here, it can no longer continue.
If N starts with 14765:
The 6th index could only be a 6, which has already been used.
If N starts with 18:
If N starts with 183:
If N starts with 18325:
2 and 8 are the only following numbers which work, and they have already been used.
If N starts with 18365:
1 is the only following number which works, and it has already been used.
If N starts with 3:
If N starts with 34:
The following number must be 5, which is already used later.
If N starts with 38:
If N starts with 381:
If N starts with 38125:
The following number must be 2 or 8, both of which have already been used.
If N starts with 38165:
N must be 3816547290.
Through casework, we can find that no other N works. Therefore, the answer is 3816547290.

• March 14 at 9:48 am

holy shit m8

• March 14 at 9:51 am

So, I started going through and your number is not evenly divis]ible by 4

• March 14 at 12:37 pm

knowing 7 is a prime number would have helped your ending a bit in that only 7 can go in the 7th place. but brute force works too! Good job!

• March 14 at 1:14 pm

Option A: 4998

• March 14 at 1:44 pm

I vote for ME at 9:09 am. That explanation was most thorough.

• March 14 at 10:53 am

• March 14 at 12:11 pm

Sorry about that, Angel was sitting on a bag of peanutz that I bought.

• March 14 at 7:19 am

Option a. 7560

• March 14 at 7:20 am

3816547290 option A

• March 14 at 9:35 am

Option B – Room 6

• March 14 at 7:20 am

22

• March 14 at 7:21 am

a: 3816547290

• March 14 at 7:21 am

Problem B: 22

• March 14 at 1:02 pm

38165444290 problem a

• March 14 at 7:23 am

Option B…3

• March 14 at 1:03 pm

3816547290 problem a

• March 14 at 7:25 am

Option A: 9786450213
Option B: 202
Option C: 1

• March 14 at 7:25 am

9735846120 for problem a

• March 14 at 8:20 am

Nicely done KC Hastings, nicely done! I was close, but seven took me awhile.

• March 14 at 1:17 pm

KC, I tried solving Option A the same way. I guess I didn’t understand the question until I read some of these answers. My answer was 9123478560. I thought I needed to solve for a 10-digit number that was divisible by numbers 2-9. My apologies. Hats off to the winners. Good eating for the next 3.14 years. Congratulations.

• March 14 at 8:51 am

and how do you figure 97358 is divisible by 5?
Option A – 3816547290
the first number is any digit [1-9] , the second must be even, the third divisible by 3, the fourth by 4, the fifth can only be 0 or 5, the sixth must be even and divisible by 3 and so on.

• March 14 at 9:33 am

You’re assuming the digit numbering goes left-to-right. I haven’t checked the math for this entry, but I suspect the opposite assumption was made.

• March 14 at 3:12 pm

Option A – 9345678120

• March 14 at 8:12 pm

C: since we need both pairs and ‘triple from triples’ it must begin after at least 3 triples and also be able to be in pairs. So, it may be any even multiple of 3 beginning at 12.

• March 14 at 7:25 am

Problem C: 64

• March 14 at 7:26 am

Option A: 3816547290

• March 14 at 7:27 am

B 314

• March 14 at 7:27 am

Problem A: 3816547290

• March 14 at 7:27 am

ANSWER for problem a is 5

• March 14 at 7:27 am

A 1234567890

B Room 11

C Multiples of 6 (6, 12, 18, 24, 30, . . .)

• March 14 at 7:28 am

Answer to Option B: any multiple of 10

• March 14 at 7:29 am

option 1: 3816549270 option 2 :0 option 3 6

• March 14 at 7:29 am

A: 7293648150

• March 14 at 7:29 am

Option A: This is a trick question if you were to leave use the numbers 1-10 tell and they were all distinct you would have an 11 digit number instead of a 10 digit number since 10 is a two digit number

• March 14 at 8:33 am

Option A: 3816547290

• March 14 at 8:52 am

It didn’t say to use 1-10, it said to use 10 distinct digits [0-9]

• March 14 at 8:53 am

It isn’t a trick question. 1,2,3,4,5,6,7,8,9,0. 0 is a digit. That leaves 10 possible digits.

• March 14 at 7:29 am

C 3

• March 14 at 7:30 am

its 3.14

• March 14 at 7:30 am

Problem A
2520000000

• March 14 at 7:30 am

For problem B the room is 314

• March 14 at 7:30 am

Option A: 9657834120
Have to start with divisibility rules. For n=1 the options are 0, 2, 4, 6, 8 since we also will need it to be divisible by 2, 5, and 10. This eliminates some options and we are left with only 0 as a possibility since it is divisible by 2, 5, and 10. Then for the 2nd digit we have different possibilities until we look at the 3rd digit. The sum of the 3 digits needs to be divisible by 3 so we start looking at the various options (similar to the start of the problem). For 4 digits we need the last 2 digits to also be divisible by 4 which narrows the possibilities again. We use the same techniques of divisibility rules as we keep adding digits.

• March 14 at 8:34 am

Option A 3816547290
Same premise as original post except this time in order from left to right.

• March 14 at 7:30 am

Option A

3816547290

• March 14 at 7:32 am

Option B= 0

• March 14 at 7:33 am

Option b =2

• March 14 at 7:35 am

Option B – 101

• March 14 at 7:35 am

Option A: 1

• March 14 at 2:24 pm

The answer to option A is 3816547290.

• March 14 at 7:38 am

Option a: 7,831,549,620

• March 14 at 7:38 am

Option A: 3,816,547,290

• March 14 at 8:22 am

Option A 3816547290

• March 14 at 7:39 am

• March 14 at 7:40 am

Option a – 3816547290

• March 14 at 7:41 am

option a is 10!

• March 14 at 7:43 am

Option b: 314

• March 14 at 7:49 am

OPTION B; 1
OPTION C: 12

• March 14 at 7:49 am

option A
9731428560

• March 14 at 10:30 am

Answer to C is 9+6n where N is any whole number greater than or equal to 1.

• March 14 at 7:50 am

Problem b: room 0

• March 14 at 10:32 am

Answer to C is 9+(6n) where n is any whole number greater than or equal to 1

• March 14 at 7:50 am

1,3,5,7,9,11,13,15,17,19

• March 14 at 7:53 am

Option B the answer is 4

• March 14 at 10:59 am

Option A: 3816547290
Option B: 2
Option C: multiples of 6

• March 14 at 7:54 am

Part A: 3816547290

What I did was since the number is divisible by 10, the last digit is 0. Then, the 5th digit must be 5. Since 1+2+…+9 is divisible by 9, the 9th digit does not matter. Note that the even digits must go in the even positions. So, the 7th digit is odd, and then the 8th digit must be 2 or 6. We try 2. _ _ _ _ 5 _ _ 2 _ 0 is our number. Then, the 6th number must be even. We try 4. Then, the sum of the first 6 digits is a multiple of 3. But, we know that the first 3 digits have a sum divisible by 3, and so do 5th+6th. Therefore, the 4th digit is divisible by 3. It is also even, so it must be 6. Now, we have _ _ _ 6 5 _ _ 2 _ 0. The digits left are 1, 3, 4, 7, 8, 9. Three of these add to a multiple of 3. Only 1 can be even. Also, the 6th7th8th is a multiple of 8, and 6th is even. So, 7th8th is divisible by 8. Then, 7th is 3 or 7. Simple casework shows we want it to be 7. More casework shows 1st2nd3rd= 381 (due to the restriction for the 7th digit).

Therefore, the number is 3816547290.

• March 14 at 10:47 am

I am not seeing that number divisible by 4 though.

• March 14 at 2:05 pm

it is indeed 3816547290

3 / 1 = 3
38 / 2 = 19
381 / 3 = 127
3816 / 4 = 954
38165 / 5 = 7633
381654 / 6 = 63609
3816547 / 7 = 545221
38165472 / 8 = 4770684
381654729 / 9 = 42406081
3816547290 / 10 = 381654729

• March 14 at 7:55 am

option B: 0

• March 14 at 8:27 am

b 13

• March 14 at 8:45 am

Option A: 3816547290

• March 14 at 9:59 am

Option B: Room 7

• March 14 at 7:55 am

A. 3816547290

• March 14 at 7:59 am

Option C : 3

• March 14 at 7:59 am

C:3

• March 14 at 8:03 am

For problem a
1827364590

• March 14 at 8:03 am

A. 3816547290

• March 14 at 8:06 am

Option b: 314
Option c: 31

• March 14 at 8:09 am

option A 3816547290

• March 14 at 8:10 am

Option A: 3816547290

• March 14 at 8:10 am

3816547290 by using a pandigital sociable group, i.e. the last number chained with the very first number

• March 14 at 8:11 am

Option c. 9 rings with 2in. Diameter

• March 14 at 8:13 am

Option C is 1234567890

• March 14 at 8:13 am

A) 3816547290
B)283
C)3.14159265358979

• March 14 at 8:13 am

• March 14 at 8:13 am

Option A — the first 9 integers are 1 -9 in any order. The last integer is 0. n = 10

• March 14 at 8:14 am

A=3816547290
C=6

• March 14 at 8:15 am

Option A= 3816547290

• March 14 at 8:15 am

9632581470

• March 14 at 8:18 am

9632581470 for option A

• March 14 at 12:32 pm

Question A:3816547290
Such a number doea not exist.

• March 14 at 2:09 pm

Your number doesn’t work for n=8 because 814 is not divisible by 8

• March 14 at 8:16 am

Option B i believe to be is 36 rooms
Option C i also believe ro be is 6

• March 14 at 8:17 am

option B 1
option C 27

• March 14 at 8:17 am

C is 7

• March 14 at 8:17 am

B i believe to be is 36 rooms
C i also believe ro be is 6

• March 14 at 8:17 am

Option A: 3816547290

• March 14 at 8:18 am

A is 3816547290

• March 14 at 8:20 am

Option A
1234567890

• March 14 at 8:20 am

B is 36
C is 6

• March 14 at 8:20 am

1

• March 14 at 8:24 am

B I beleve is 36
C I believe is 6

• March 14 at 8:24 am

Option C- Multiples of 3

• March 14 at 8:24 am

Option C 6

• March 14 at 8:25 am

A) 1834567290
B) 10
C) multiples of 7

• March 14 at 8:37 am

C: 12 +6n

• March 14 at 8:25 am

Option A: The number is 1234567890

• March 14 at 8:26 am

Option A: 3816547290

• March 14 at 8:27 am

OPTION A: 9,876,543,210

• March 14 at 8:28 am

3816547290 problem A(option A)

• March 14 at 8:30 am

Digit: 1 2 3 4 5 6 7 8 9 10
Value: 1/3/7/9 2/4/6/8 1/3/7/9 2/4/6/8 5 2/4/6/8 1/3/7/9 2/4/6/8 1/3/7/9 0

• March 14 at 8:28 am

Option A: 3816547290

• March 14 at 8:29 am

Option A – 9,876,351,240

• March 14 at 8:31 am

1252566450

• March 14 at 8:31 am

A 4
B 1
C 3

• March 14 at 8:31 am

For option B, the number on the door is 13.

• March 14 at 8:32 am

Option A
1032547698
N=2

• March 14 at 8:33 am

Option A
1,111,111,111

• March 14 at 8:34 am

Option A: 3816547290

• March 14 at 8:35 am

1264564800 for Option A

• March 14 at 8:36 am

Option A: 1234567890

• March 14 at 8:37 am

http://puzzling.stackexchange.com/questions/3017/10-digit-number-where-first-n-digits-are-divisible-by-n

Very original. All you can do in <1min of page going live is to Google it.

How hare would it have been to google your own damn problem statement, Conway?

Give it to the next guy who actually coded it or worked it out.

• March 14 at 9:10 am

How do you know what I did? I’d done that problem like 10 years ago. I read it in a book and solved it myself. I even proved that it was the unique answer. So yeah the problem was unoriginal, and yeah I didn’t solve it on the spot, but that’s not my problem. I answered it first because I remembered the answer, not because I looked it up.

• March 14 at 10:30 am

Option A
3816547290

• March 14 at 8:37 am

This might sound silly but where do u comment or how do u comment your answer

• March 14 at 8:38 am

Well done.

• March 14 at 8:41 am

Option A: 3,816,547,290

• March 14 at 8:42 am

problem A: 3816547290

• March 14 at 8:43 am

• March 14 at 8:44 am

Option A: 9657834120

• March 14 at 8:46 am

Option C: i think 9, 15, 21, 27, 33….

• March 14 at 8:46 am

option a

321654987

• March 14 at 8:47 am

• March 14 at 8:49 am

Option A=1

• March 14 at 8:50 am

A: 3816547290
B: 10
C: multiples of 7

• March 14 at 8:50 am

One answer to A is 3816547290.

• March 14 at 8:51 am

For option A: 3816547290

• March 14 at 8:52 am

Ooption A=1

• March 14 at 8:53 am

3816547290 for A.

• March 14 at 8:54 am

Option A…..1234567890

• March 14 at 8:55 am

A. 1234759680 c. 12

• March 14 at 8:55 am

Option A: 3816547290

• March 14 at 8:56 am

option a 2468109573

• March 14 at 8:56 am

Problem B : 6

• March 14 at 8:56 am

option A

Option A=1

• March 14 at 8:57 am

Option A = 3816547290

• March 14 at 8:58 am

Option A is 3816547290

• March 14 at 8:59 am

Option C: Multiples of 9

• March 14 at 8:59 am

B.) 1

• March 14 at 9:01 am

Option A = 3816547290

Option C:
Any multiple of 6.
I.e. 6, 12, 18, …..

• March 14 at 9:02 am

Option C is 3, if the links are intertwined.

• March 14 at 9:03 am

Option A: 3816547290

• March 14 at 9:03 am

For option a the answer is 2520

• March 14 at 9:06 am

Option a
3816547290

• March 14 at 9:07 am

Option A: 120456789

• March 14 at 9:08 am

One fairly obvious answer to question B is the perfect number 6.

• March 14 at 9:08 am

The answer to problem A is 0123456789

• March 14 at 9:08 am

• March 14 at 9:10 am

• March 14 at 9:10 am

0123456789 for option a

• March 14 at 9:10 am

• March 14 at 9:10 am

• March 14 at 9:11 am

Option b: room 0

• March 14 at 9:11 am

Option A n=6
BECAUSE
317520/6=52920
317520/12=26460
317520/18=17640
317520/24=13230
317520/30=10584
317520/36=8820
317520/42=7560
317520/48=6615
317520/54=5880
317520/60=5292

Therefore n=6 when
Number=317520

• March 14 at 9:12 am

Option c: 33

• March 14 at 9:13 am

Opton A 3816547290

• March 14 at 9:15 am

Option B is 55

• March 14 at 9:16 am

3816547290 for A

• March 14 at 9:16 am

The answer to problem A is 1234567890
The answer to problem B is 6.
The answer to problem C is 5

• March 14 at 9:17 am

option b : room zero

• March 14 at 9:17 am

Option A is:
3816547290
Option B is:
7
Option C is:
27

• March 14 at 9:19 am

Option A: 3816547290

• March 14 at 9:20 am

Option A:3816547290

long i,j,n,c; for (i=1000000080; i1; n/=10,j–) if (n%j) break; if (j==1) {c++; printf(“%ld “,i);} }} printf(“\nTotal: %ld\n”,c); Only integer with all numbers 0-9.

• March 14 at 9:20 am

Option A: 5312948760

• March 14 at 9:20 am

Option A is 314

• March 14 at 9:23 am

This sounds like topology or knot theory, neither of which I have studied, so my guess will be 1 key chain with three rings linked together. This is because there is no left or right handedness in addition to not being able to distinguish between any pair of key rings. This was a lot of fun. Thanks!

• March 14 at 10:05 pm

The answer is all possible total number of rings, not just one number. Cannot dif. Between pairs, must be odd. B/c it can’t be a multiple of 2. However, you can dif. Between triples, so it must be any multiple of 3 that is not also a multiple of 2. B/c it says triple from triples there must be more than 1 triple therefore the answer cannot be 3. So, the answer (A) is equal to 3x, when x is odd and x>1. So the solution would be 9,15, 21, etc. This would also make it where you could not dif. Between left and right sides.

• March 14 at 9:24 am

Option C: 32

• March 14 at 1:00 pm

Option a is 3816547290

• March 14 at 9:25 am

• March 14 at 9:26 am

Option A
1234759680

• March 14 at 9:26 am

Option C is 9. 2 variations possible with triples: linear or ring. 2 linear triples connected with ring triple on end, or 2 linear triples with ring triple on end (=9 rings total) would allow differentiation of each of three triple sets without knowing left from right by position: middle, ring triple end, or linear triple end.

• March 14 at 9:26 am

Option A:
-10 digits that are all distinct is 0-9
– Let n=5 –> the first 5 of them is divisible by 5 and 10 (ends in 5 or 0)
– Every 5 numbers should be divisible by 5 and 10
–> number is 1234567890
–maybe-ish???

• March 14 at 9:33 am

Nope. As per usual my mistake was in the English, not the numbers.
Wow so simple and I was totally going the wrong way.
n=2, first 2 numbers divisible by 2
n=3, first 3 numbers divisible by 3

n=9, first 9 numbers divisible by 9
You got me.

• March 14 at 9:27 am

Option A: 6234513789

• March 14 at 9:28 am

Option A: 3816547290

• March 14 at 9:29 am

Ooops, looks like no free pizza at our house.

• March 14 at 9:29 am

My answer is for Option B . The number of the room is 13.

• March 14 at 9:30 am

A: 3816547290

• March 14 at 9:30 am

A:3816547290
B: 4
C: 3

• March 14 at 9:31 am

The answer for option A is 87687687.

• March 14 at 9:31 am

Option B: 310

• March 14 at 9:32 am

Option a cant be solved because no number is divisible by 0.

• March 14 at 9:32 am

A is. 3816547290

• March 14 at 9:32 am

Option C: 9

• March 14 at 9:33 am

Option C: odd numbers beginning at 5 (ex. : 5,7,9…)

• March 14 at 9:34 am

Classroom # 8

• March 14 at 9:34 am

Option A is: 9,523,478,160

• March 14 at 9:34 am

Option C: perfect squares that are divisible by 3 (9,16,36,81….)

• March 14 at 9:35 am

option B 102

• March 14 at 9:36 am

Option A:
9, 356, 712, 480

• March 14 at 9:36 am

the answer to A is 3816547290

• March 14 at 9:38 am

Problem A

1023456789

• March 14 at 9:39 am

Option C – 5 rings

• March 14 at 12:25 pm

A. 3816547290

B. Room 1

C. 5 Rings

• March 14 at 9:40 am

Option A: the answer is 3816547290 as many have already posted. But the interesting thing about this answer is that if you carry pi out to enough digits you will actually find the number sequence 3816547290 within it!

• March 14 at 10:28 am

Yep! According to http://www.subidiom.com/pi/pi.asp, “the numeric string 3816547290 appears at the 1,481,722,654th decimal digit of Pi.”

• March 14 at 9:40 am

Option A = 9657834120

• March 14 at 9:42 am

Option A: 1234567890

• March 14 at 9:42 am

Option B 1 or room 1

• March 14 at 9:43 am

1- 3816547290
2-314
3- 3.14159265…..

• March 14 at 9:44 am

Option A – 3816547290

• March 14 at 9:46 am

3816547290

• March 14 at 9:47 am

A: 3816547290

• March 14 at 9:48 am

Option c 3.141592

• March 14 at 9:49 am

3816547920 option a

• March 14 at 9:48 am

Option A: 987654321

• March 14 at 9:50 am

Here is a small perl program.

sub pls_check {
local (\$ar_ref, \$dig) = @_;

if (\$dig == 1) {
for \$i (1..9) {
push (@\$ar_ref, \$i);
}
return;
}

local @ar_dig_minus_1 = ();
pls_check(\@ar_dig_minus_1, \$dig – 1);
foreach \$i (@ar_dig_minus_1) {
for \$j (0..9) {
my \$j_str = “j”;
if (“\$i” =~ m/\$j/) { next; }
my \$n = \$i * 10 + \$j;
if (\$n % \$dig == 0) {
push (@\$ar_ref, \$n);
if (\$plg_stop) {return;}
}
if (\$plg_stop) {return;}
}
if (\$plg_stop) {return;}
}
return;
}
@ar_n = ();
\$n = 10;
pls_check(\@ar_n, \$n);

• March 14 at 10:27 am

wrong

• March 14 at 9:51 am

• March 14 at 9:51 am

A.
3816547290

• March 14 at 1:07 pm

Option C: π itself.

• March 14 at 9:51 am

Option A: 3816547290

• March 14 at 9:52 am

Option A: 9735846120

• March 14 at 9:52 am

Answer to question no. 1 is 1111111111

• March 14 at 9:55 am

3816547290 for problem A

• March 14 at 9:55 am

Option A:
1,4,9,16,25,36,49,64,81,100

• March 14 at 9:55 am

One for option B

• March 14 at 9:56 am

Option B: 12

• March 14 at 9:56 am

Option c is 3 rings

• March 14 at 9:56 am

0

• March 14 at 9:56 am

B: room 1

• March 14 at 9:58 am

1987654320

• March 14 at 9:58 am

A: 3816547290, but no elegant way to find it other than brute force and confirmation here.

• March 14 at 9:59 am

Option A: 3816547290
B. 314
C. 3.1415926535897932……… infinity

• March 14 at 9:59 am

Option A is 3816547290.

• March 14 at 5:35 pm

Option A: 1
Option B: 1
Option C: 5

• March 14 at 10:00 am

3816547290

• March 14 at 10:03 am

Option B: 9

• March 14 at 10:04 am

Option B is 120

• March 14 at 10:05 am

3816547290

• March 14 at 10:05 am

Option A=3816547290
Option B=314
Option C=3.14159265358979323846…..

• March 14 at 10:06 am

Option C: 9

• March 14 at 10:06 am

Option A: 3816547290

• March 14 at 10:06 am

1111111111
3
8

• March 14 at 10:07 am

Option B: 101

• March 14 at 10:08 am

Option A is 3816547290

• March 14 at 10:08 am

Problem A: 3,816,547,290

• March 14 at 10:09 am

Option b is 100

• March 14 at 10:10 am

Problem B Room 101…..the numbers are 100,1 and 0

• March 14 at 10:11 am

For option A I got 9876543210

• March 14 at 10:11 am

• March 14 at 10:11 am

Option C is 27.

• March 14 at 10:13 am

3816547290 for question A

• March 14 at 10:13 am

Option C
10

• March 14 at 10:13 am

Option A = 1234567890

Option B = 0

Option C = 6

• March 14 at 10:14 am

Option A=3816547290
Option B=314
Option C=3.141592653589793238462643383279502884197169399375105820974944592307

at least I think these are the right answers …

• March 14 at 10:15 am

Option 2: 8

• March 14 at 10:16 am

Option A: 3816547290

• March 14 at 10:16 am

Option B = 45

• March 14 at 10:17 am

A) 1,234,759,680

• March 14 at 10:17 am

question #1 123487560

• March 14 at 10:17 am

Option A: 3816547290

• March 14 at 10:18 am

sorry….. 1234875609

• March 14 at 10:18 am

Option a
3816547290

• March 14 at 10:19 am

For Option/Problem A
3691287540

• March 14 at 10:20 am

Option A: 3816547290
Option B:109
OptionC:15

• March 14 at 10:20 am

Option A 3816547290

• March 14 at 10:22 am

Option B: 14

• March 14 at 10:22 am

Option b: 21

• March 14 at 10:24 am

Option C: 72

• March 14 at 10:25 am

• March 14 at 10:25 am

Option A.) 3816547290

• March 14 at 10:26 am

A: 3816547290
B: 4
C: 3.1415926535

• March 14 at 10:26 am

Option B
4

• March 14 at 10:27 am

Option A: 3816547290

• March 14 at 10:29 am

Option c: any odd multiple of 3
(3,9,15,21, etc.)

• March 14 at 10:29 am

A: 3816547290

• March 14 at 10:31 am

Problem A: 3816547290

• March 14 at 10:31 am

really oh my gosh iam in college that is so easy

• March 14 at 10:31 am

Question A: 3816547290. 381 is divisible by 3, 38165 is divisible by 5, and so forth.

• March 14 at 10:31 am

Option A=3816547290

Option B=314

Option C=3.141592653589793238462643383279502884197169399375105820974944592307

• March 14 at 10:31 am

Problem A 3816547290

• March 14 at 10:33 am

The room number is 314

• March 14 at 10:33 am

B: 4 or 5

• March 14 at 10:33 am

A is 3890264839
B is trick question. Cannot tell order from a communitive property since multiplication is communitive the product of 287 is the same as 827
C is 6.

• March 14 at 10:34 am

C is 1023456789

• March 14 at 10:34 am

Option C: One

• March 14 at 10:34 am

Option A: 3816547290

• March 14 at 10:35 am

option A= 3816547290
option B= 314
option C= 11

• March 14 at 10:35 am

C=9

• March 14 at 10:36 am

3816547290 that is for problem A.

• March 14 at 10:39 am

Option C: 6

• March 14 at 10:39 am

Option A = 3816547290 Is my answer

• March 14 at 10:40 am

3816547290

• March 14 at 10:40 am

Problem A: 3816547290

• March 14 at 10:41 am

A: 1,234,759,680
B: pi
C; 27

• March 14 at 10:42 am

• March 14 at 10:42 am

Option A to pie contest
2496543210

• March 14 at 10:43 am

option c 9

• March 14 at 10:44 am

Option B: The room number is 12
(Assuming the numbers in the list are all positive integers)

• March 14 at 10:45 am

option a – 1357111317

option b – 1

• March 14 at 10:46 am

For option b the answer is room 4, the product of two numbers (2) is the same as the addition of them.

• March 14 at 10:48 am

Option A also has 5698743120 as a solution

• March 14 at 10:50 am

Option A: 3816547290

• March 14 at 10:52 am

Option B: The room number is equal to the product + 1 where the product(calculated by the girl) is a prime number.

• March 14 at 10:53 am

A) 3816547290

• March 14 at 10:53 am

Option B: room 0

• March 14 at 10:54 am

Option A: 8,697,543,120

• March 14 at 10:55 am

Option A is 8,697,543,120

• March 14 at 10:55 am

Option B: 1

• March 14 at 10:56 am

Answer to C: Every other multiple of three, starting at six…because these numbers are all divisible by both three and two, thus allowing them to have hard to distinguish pairs and much more noticeable triples without leaving one key ring on its own.

• March 14 at 10:56 am

For option A:
1234567890

• March 14 at 10:57 am

A) 9657834120

• March 14 at 10:57 am

3816547290 for option A

• March 14 at 10:58 am

A=3816547290
B=3.14
C=3.14159265389

• March 14 at 11:00 am

Option B: 314
Option C:3.14159265359

• March 14 at 11:00 am

are you sure i don’t know

• March 14 at 11:01 am

Just a bonus problem to think about. That much pizza would equate to 3.3million calories or 940lbs

• March 14 at 3:32 pm

Option c
Pair = 2
Triple=3
L/r=2
2*3*2=12
All multiples of 12….12, 24, 36, 48…

• March 14 at 11:02 am

• March 14 at 11:03 am

Option A: 3.1415926535

• March 14 at 11:04 am

• March 14 at 11:05 am

Option A:
3,216,549,870

• March 14 at 11:06 am

3816547290
For option A

• March 14 at 11:06 am

A: 9.86460093227
B: 9.86460093227
C: 9.86460093227

• March 14 at 11:07 am

Option B = 102

• March 14 at 11:08 am

Option B:
6

• March 14 at 11:09 am

B-314

• March 14 at 11:11 am

• March 14 at 11:11 am

Option B = 4

• March 14 at 11:12 am

Option A: There is no unique solution to this problem as it is never specified that the quotient of dividing the number by n result in an integer. For example, the answer very well could be the 1234567890.

To further clarify

1234567/7=176366.714285714, this is not an integer but obviously 1234567 is divisible by 7.

• March 14 at 1:24 pm

Nope. In mathematics, we say that an integer n is divisible by k if and only if n can be written as the product of k and q, where q is an integer.

• March 14 at 11:12 am

My guess for Option 1. 2,520,252,000

• March 14 at 11:12 am

Option A : 2013456789

• March 14 at 11:13 am

for A= 3816547290

• March 14 at 11:13 am

option A: 3816547290

• March 14 at 11:13 am

Option A: 120456045

• March 14 at 11:14 am

option A 3816547290

• March 14 at 11:15 am

Option A – 3816547290

• March 14 at 11:17 am

Option b is room 001

• March 14 at 11:17 am

for A i got 0.31

• March 14 at 11:18 am

A) 3816547290

• March 14 at 11:19 am

Option A is 3816547290

• March 14 at 11:19 am

• March 14 at 11:19 am

Option c is 15

• March 14 at 11:20 am

Option A – 9

• March 14 at 11:21 am

THE ANSWER TO OPTION a IS 1836547290

• March 14 at 11:22 am

Option “a” – 3816547290
Option “b”- 0
Option “c”- 6

• March 14 at 11:22 am

• March 14 at 11:23 am

1234567890

• March 14 at 11:23 am

option c is π

• March 14 at 11:24 am

Option A: 3816547290

• March 14 at 11:24 am

Option a is 3816547290
Option b is 314
Option c is 3.1415926535897932

• March 14 at 11:25 am

The answer to Question A is 3816547290

• March 14 at 11:26 am

Option A is
I think
1232561610

• March 14 at 11:26 am

A- 1234759680

• March 14 at 11:27 am

I’m really late, but I did the work so I’m going to post anyway.

Option A: 3816547290
Option B: Room 0
Option C: odd numbers divisible by 3

• March 14 at 11:28 am

Option A:3816547290

• March 14 at 11:29 am

Option c: 4 and up. the 4 ring solution has these rings connected: a-b, b-c, c-d, d-b. You can tell triple b, c, d from all other triples. A strange jumble could be a large loop, a-b, b-c, c-d… y-z, z-a plus one additional ring, alpha. alpha-a, alpha-b. You can tell triple alpha, a, b apart from any other triple.

• March 14 at 11:29 am

Option A: 4
Option B: 1
Option C: 3

• March 14 at 11:30 am

• March 14 at 11:30 am

Problem A) 3816547290
Problem B) 314
Problem C) 6 (multiples)

• March 14 at 11:31 am

Question A: 3816547290
Question B: 314
Question C: 3.14

• March 14 at 11:31 am

Problem A: the answer is 3816547290

• March 14 at 11:32 am

3

• March 14 at 11:32 am

problem c : 3

• March 14 at 11:33 am

• March 14 at 11:34 am

C. 6

• March 14 at 11:35 am

Problem A:
3816547290
Problem B:
314
Problem C:
6

• March 14 at 11:36 am

3816547290

• March 14 at 11:36 am

3816547290 but i know it has been posted a bunch of times already

• March 14 at 11:36 am

Problem A: 9817524360

• March 14 at 12:16 pm

Problem A correction: 9817524360

• March 14 at 12:49 pm

Final Correction: for Problem A: 9417825360

• March 14 at 4:13 pm

Explanation: My approach has the digits numbered from right to left (not left to right as the other answers seem to follow). Since the question does not specify which direction the digits are ordered, I reasoned that the digits should be numbered from Right to Left following the ones, tens hundreds, thousands, etc places.

• March 14 at 11:36 am

Question B

• March 14 at 11:37 am

• March 14 at 11:38 am

MATH IS NOT REAL.

WE ARE NOT REAL.

WHAT IS LIFE?

• March 14 at 11:38 am

Option A = 3816547290

• March 14 at 11:39 am

option c: 7

• March 14 at 11:39 am

Option B: 32

• March 14 at 11:39 am

Option A = 3 816 547 290

• March 14 at 11:40 am

For Option A, answer is 1834567290

• March 14 at 11:40 am

• March 14 at 11:41 am

Option C= 3

• March 14 at 11:41 am

• March 14 at 11:41 am

The answer for question A is 3816547290

The answer to question B is room 314

The answer to question C is 9 +6N where N is greater than or equal to 1

• March 14 at 11:42 am

Option C: any multiple of 2 starting with 6

• March 14 at 11:43 am

Option A: 50967832

• March 14 at 11:43 am

Option A: 5096783214

• March 14 at 11:43 am

3816547290 – Problem A

• March 14 at 11:44 am

Option A; 9274561830
Option C; 3+3n, where n is any integer

• March 14 at 11:46 am

Option A: 3816547290

• March 14 at 11:47 am

A. 3816547290

B. 314

C. 3.141592654

• March 14 at 11:49 am

For C, You never said the answer had to be in numerical format…… I call the ten digit integer a phone number…. Who do you call for pizza? Pizza Hut. the Area Code and the local number to pizza hut…. I have that number on my fridge. Could never forget…
It could also be an 1800 number… I know, pretty clever. #RulesOfAnswer (Here’s to charm) I’m a fan, not a mathmetician

• March 14 at 11:50 am

Option C: 3,9,27, or 81.

• March 14 at 11:52 am

3816547290

• March 14 at 11:52 am

• March 14 at 11:52 am

Option A=3816547290
Option B=21
Option C= 5

• March 14 at 11:52 am

• March 14 at 11:52 am

problem A: 3816547290

• March 14 at 11:53 am

Option A: 3816547290

• March 14 at 11:53 am

Option B is 312

• March 14 at 11:55 am

Option C: 7

• March 14 at 11:57 am

A is 3816547290

• March 14 at 11:57 am

The answer to the problem B is 36.

• March 14 at 11:58 am

• March 14 at 11:58 am

Option A: 1254763890 n=2

bam

• March 14 at 11:58 am

Option B room # is 314

• March 14 at 11:58 am

If the first n digits works from right to left, rather than left to right, then Option A is 9687435120.

• March 14 at 11:58 am

• March 14 at 12:01 pm

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

• March 14 at 12:01 pm

Option A: 3816547290

• March 14 at 12:01 pm

A . 3816547290
B . 314
C . 3.141592653589

• March 14 at 12:02 pm

Option A . 3816547290
Option B . 314
Option C . 3.141592653589

• March 14 at 12:02 pm

A: 3816547290

• March 14 at 12:02 pm

Option C= pie

• March 14 at 12:02 pm

• March 14 at 12:03 pm

A: 3816547290

• March 14 at 12:03 pm

3816547290 for A

• March 14 at 12:03 pm

Option B:

Room 0

• March 14 at 12:06 pm

Option A=3816547290
Option B=21
Option C= 5

• March 14 at 12:06 pm

A) 3816547290
B) 314
C) 3.1415926535897932

• March 14 at 12:07 pm

the answer to problem A is 1296547830

• March 14 at 12:07 pm

This isn’t actually a reply, but my submission. I couldn’t find the comment button on my phone. My answers are a. 3816547290, the only possible answer I can come up given the info. B. Is any room number that has a 0 in it. C. Is 3×2n, where n is any positive real number.

• March 14 at 12:07 pm

OPTION A: 3216748950

• March 14 at 12:07 pm

Option B: Not solvable with information given.

The question says “numbers.” It says nothing about whether they are positive or negative, integral or decimal, rational or irrational, or even if they are distinct. Zero could also be included. The same applies for the room number.

As the question is stated, there are an infinite number of potential answers.

• March 14 at 12:09 pm

3816547290

• March 14 at 12:10 pm

A:3816547290

• March 14 at 12:10 pm

Option a has 3 correct answers. 1834567290. 3816547290. And 9814567230.

• March 14 at 12:11 pm

Option A: 0123456789

• March 14 at 12:12 pm

3816547290 for problem A

• March 14 at 12:13 pm

C: 6,12,18,24

• March 14 at 12:13 pm

I Say option A: is go fuck yourself
Option B: is i lit your baby on fire
Option C: this is a setup

• March 14 at 12:14 pm

option B= room 314

• March 14 at 12:14 pm

option A 3816547290

• March 14 at 12:14 pm

The FIRST n for Option A is : 1234568160

• March 14 at 12:21 pm

Oops… copied the wrong value: Option A: 1234759680

• March 14 at 12:15 pm

Option A…..1234567890

• March 14 at 12:17 pm

YOUR PIZZA’S NOT THAT GOOD TO TRY TO DO THIS

• March 14 at 12:18 pm

Option B – 0

• March 14 at 12:18 pm

option C:5

• March 14 at 12:19 pm

Option A: 3816547290

• March 14 at 12:19 pm

Option A:
3816547290

• March 14 at 12:19 pm

A: 3816547290

• March 14 at 12:19 pm

Option A=3816547290
Option B=314
Option C=3.141592653589793238462643383279502884197169399375105820974944592307

• March 14 at 12:20 pm

Option A is 3816547290

• March 14 at 12:20 pm

3816547290 for option A

• March 14 at 12:21 pm

A. 987, 456, 3210

• March 14 at 12:21 pm

• March 14 at 12:22 pm

For Option A: 1234567890

• March 14 at 12:22 pm

A is 2520

• March 14 at 12:22 pm

3816547290 for Problem A.

• March 14 at 12:22 pm

• March 14 at 12:22 pm

Problem B: 314

• March 14 at 12:23 pm

problem 1: 1234567890

• March 14 at 12:23 pm

Option C: 4.

• March 14 at 12:24 pm

A) 3816547290
B) 23
C) 5,9,13,… Or anything that takes the form of the function f(x) = 4x + 5, where x must be an interger > or = 0

• March 14 at 12:24 pm

Option A: 3816547290

• March 14 at 12:24 pm

Option A: 9,738,561,240

• March 14 at 12:24 pm

Option A: 3816547290

• March 14 at 12:25 pm

1112222520 for option c

• March 14 at 12:26 pm

A correct answer for A is 1120120120.
2 divides 20
3 divides 120
4 divides 120
5 divides 20120 (because it ends in 0)
6 divides 120120 (6 times 20020)
7 divides (0120120) (because 7 divides 1001 and 1001(120)=0120120)
8 divides (20120120) (because 8 divides both 1000 and 120)
9 divides (120120120) (because the digits sum to 9)
10 divides (1120120120) (because the last digit is 0)
This answer is not unique certainly any number can be used for the highest place digit and the two lower digits only need to have a sum that is 3 mod 9. There are probably many more answers than that.

• March 14 at 12:26 pm

Option A:
3,816,547,290

• March 14 at 12:27 pm

1234567890

• March 14 at 12:28 pm

option c:

9, 12, … (n+2)3 where n is any real integer greater than zero

• March 14 at 12:29 pm

Option C

• March 14 at 12:29 pm

Question A: 3816547290

• March 14 at 8:57 pm

2 sets of pairs =8
3 sets of triples=9
(math never my strong point lol)

• March 14 at 12:29 pm

A. 3816547290

• March 14 at 12:30 pm

Looks like everyone and there mother has already posted it, presumably because it’s easy to check once someone else has, but anyway: Option A is 3816547290. I managed to get it myself though. My solution wasn’t very elegant though: it basically amounted to getting it down to 10 or so cases and then checking the first 7 were divisible by 7 by hand. I wonder if there’s a better way.

• March 14 at 12:30 pm

Option A: 1,472,583,690 is another combination right?

• March 14 at 12:30 pm

Option B: 22

• March 14 at 12:31 pm

Oops I read Option A wrong.

• March 14 at 12:32 pm

Question A: 3816547290
Question B: 314
Question C: 3.14

• March 14 at 12:34 pm

3816547290 Option A

• March 14 at 12:34 pm

option a
9147528360

• March 14 at 12:34 pm

option a= 1234567890

• March 14 at 12:34 pm

• March 14 at 12:36 pm

DANK MEMES

• March 14 at 12:36 pm

OPTION A
123456789
N=1

• March 14 at 12:37 pm

Option A: 1584923760

• March 14 at 12:37 pm

3816547290 for Problem A

• March 14 at 12:39 pm

My answer to Question B is Room 314

• March 14 at 12:39 pm

A: a very big number
B: somewhere in between 1 and 10,000
C: 4, obviously

• March 14 at 12:39 pm

Option A: 3186547290

• March 14 at 12:39 pm

Option c: Any multiple of three greater than or equal to 6

• March 14 at 12:39 pm

The answer for Option A is 3816547290

• March 14 at 12:40 pm

Answer to option B is room 314

• March 14 at 12:40 pm

Problem A:
3816547290

• March 14 at 12:40 pm

For problem A, either of these work: 1472583960, 3816547290

• March 14 at 12:41 pm

its 69 f44g0t

• March 14 at 12:42 pm

Problem A solution 9876543210 the first five is divisible by 5 as are the next five. N=5

• March 14 at 12:43 pm

• March 14 at 12:44 pm

A: 3
B:1
C:4

• March 14 at 12:45 pm

• March 14 at 12:45 pm

Option A: 3816547290

• March 14 at 12:45 pm

Option A:3816547290

• March 14 at 12:45 pm

Option A: 9,738,561,240

• March 14 at 12:47 pm

Option A: 9,876,543,040

• March 14 at 12:47 pm

Option A: 3816547290

• March 14 at 12:49 pm

Option A: 3816547290
Option c: is pie which is 3.14

• March 14 at 12:50 pm

Option A: 3816547290

• March 14 at 12:50 pm

Option B: 8

• March 14 at 12:52 pm

Option A: 3816547290

• March 14 at 12:52 pm

Option A

• March 14 at 12:54 pm

Option A
3816547290

• March 14 at 12:55 pm

Option C: 10

• March 14 at 12:56 pm

B is .413
c is. 763534687.e376300

• March 14 at 12:57 pm

Option a is 3816547290

• March 14 at 12:57 pm

3816547290

• March 14 at 12:58 pm

Option A: 3816547290

• March 14 at 1:00 pm

Lame…west coast people don’t even get a chance. 😉 A: (without looking at replies!) 3816547290. Oh well, sleeping in is fun too. 😉

• March 14 at 1:00 pm

Is 18

• March 14 at 1:00 pm

Option A is 1111111111

• March 14 at 1:00 pm

For option A: 3816547290 is the answer

• March 14 at 1:01 pm

Question A: 3816547290
Question B: 314
Question C: 3.14

• March 14 at 1:03 pm

option C: 11

• March 14 at 1:03 pm

• March 14 at 1:05 pm

Option C is; 3.14159… Pi

• March 14 at 1:05 pm

Option C is 24, 33, 69, 239, 3699, 9636

• March 14 at 1:06 pm

A 3816547290
B room 0
C 5

• March 14 at 1:07 pm

Option A: 1236543210

Thanks!

• March 14 at 1:09 pm

option A = 3941578620

• March 14 at 1:11 pm

Option A solution:
7,896,541,230

• March 14 at 1:12 pm

Here’s the solution to Option B:

Let S be a finite set of Natural Numbers (since they can’t be in a negative room number of room 0), whose cardinality is greater than 1 (from the wording of the question).

Since we know that the product of all the elements of set n, and its cardinality, the case to which the student would be correct in guessing the room number, given only their products and cardinality, is if the product of all elements in the set were also equal to the sum of all elements in the set, given that the cardinality of the set is respected.

For example, let’s say that the cardinality of the set is 3, and the product of all elements in the set equals 46. There is no possible way to represent 46 as the sum of three numbers and the product of three numbers. (46 x 1 x 1 = 46, 46 + 1 + 1 = 48).

The only case to which the sum of n number of elements, all of which are natural numbers, is equal to the product of n number of elements is 4.

They meet in room #4.

• March 14 at 1:13 pm

Option A: 3816547290

• March 14 at 1:13 pm

• March 14 at 1:13 pm

Option A: 9876543210

• March 14 at 1:14 pm

• March 14 at 1:16 pm

Option A:
1234567890

Option C:
24

• March 14 at 1:16 pm

Option A:
1234567890

• March 14 at 1:17 pm

Option c 12

• March 14 at 1:18 pm

A: 3816547290
B: 12
C: 4

• March 14 at 1:18 pm

The answers to all the questions are
A: 3816547290
B: 314
C: 3.1415926535897932…

• March 14 at 1:18 pm

Option A: 3816547290

• March 14 at 1:19 pm

Option A: 3816547290

• March 14 at 1:22 pm

[Verse 1]
Hello, it’s me
I was wondering if after all these years
You’d like to meet, to go over everything
They say that time’s supposed to heal ya
But I ain’t done much healing
Hello, can you hear me?
I’m in California dreaming about who we used to be
When we were younger and free
I’ve forgotten how it felt before the world fell at our feet
There’s such a difference between us
And a million miles

[Chorus]
Hello from the other side
I must’ve called a thousand times
To tell you I’m sorry, for everything that I’ve done
But when I call you never seem to be home
Hello from the outside
At least I can say that I’ve tried
To tell you I’m sorry, for breaking your heart
But it don’t matter, it clearly doesn’t tear you apart anymore

[Verse 2]
Hello, how are you?
It’s so typical of me to talk about myself, I’m sorry
I hope that you’re well
Did you ever make it out of that town
Where nothing ever happened?
It’s no secret
That the both of us are running out of time

[Chorus]
So hello from the other side
I must’ve called a thousand times
To tell you I’m sorry, for everything that I’ve done
But when I call you never seem to be home
Hello from the outside
At least I can say that I’ve tried
To tell you I’m sorry, for breaking your heart
But it don’t matter, it clearly doesn’t tear you apart anymore

[Bridge]
Ooooohh, anymore
Ooooohh, anymore
Ooooohh, anymore
Anymore

[Chorus]
Hello from the other side
I must’ve called a thousand times
To tell you I’m sorry, for everything that I’ve done
But when I call you never seem to be home
Hello from the outside
At least I can say that I’ve tried
To tell you I’m sorry, for breaking your heart
But it don’t matter, it clearly doesn’t tear you apart anymore

• March 14 at 1:22 pm

• March 14 at 1:22 pm

• March 14 at 1:24 pm

C – 3.141592653589793238462643383279502884197169399375105820974944592307

• March 14 at 1:25 pm

Option A: Most everyone is assuming the digits are read left to right. I took the problem to be read from right to left. Therefore my verified answer is:

Option A: 9,753,126,480.

• March 14 at 1:25 pm

• March 14 at 1:26 pm

I like number 4 for option A

• March 14 at 1:26 pm

Option C: the amount of keys has to be a number divisible by 3

• March 14 at 1:27 pm

Option A: 5678913240

• March 14 at 1:28 pm

Option a is 3816547290
Option b is 314
Option c is 3.1415926535897932

• March 14 at 1:29 pm

Option C: The answer is the multiples of 3, but since it says “I can tell some triple from other triples” we have to start with 6 rings on the chain and then all other multiples of 3 are possible numbers of key-rings in this jumble. (i.e. 6,9,12,15,18,21,24,27,30,33,36,etc.).

• March 14 at 1:29 pm

Option B the answer is 6.

• March 14 at 1:29 pm

Maybe I am loosing it, but wouldn’t a possible answer to Option A be 1296547830

• March 14 at 1:30 pm

option A is not unique, but the number you have in mind is: 9876543210
9876543210 ends in 0 (/10)
987654321 digits add to 45 (/9)
98765432 432/8 = 54 (/8)
9879643/7 = 711379 (/7)
987654 even and digits add to 39 (/6)
98765 ends in 5 (/5)
9876 76/4=19 (/4)
987 9+8+7=24 2+4=6 (/3)
98 even (/2)
9 any (/1)

do I have to post all three answers?

• March 14 at 1:31 pm

Option A: 4398765120

• March 14 at 1:31 pm

Option A = 3816547290

• March 14 at 1:32 pm

For Option A, I got 9,375,261,480.

• March 14 at 1:32 pm

The answer is 3816547290 for option a

• March 14 at 1:33 pm

Option A: 3816547290

• March 14 at 1:35 pm

Problem A: 3816547290

• March 14 at 1:35 pm

Part A: The number for part a) is 1236067290. I determined the answer by starting with an arbitrary digit from the left, 1. 1 is divisible by 1. Then add a digit to the right to have the first two digits divisible by 2. So we can use an algorithm as follows: 1) On the nth digit, divide the number formed by n+1 and find the remainder. 2) To determine the next digit, simply add a digit that will make the remainder*10 + (added digit) divisible by n+1 3) If the remainder is 0 at the nth digit, add ‘n+1’ to the right. For example, 123606/7 = 0, so add ‘7’ to the right to form 1236067. Then, 1236067/8 = 154508 remainder 3. We know that 3*10+2 = 32 is divisible by 8, so the next digit is 2. We continue this process until we reach the tenth digit. It is known that if a number is divisble by 10, its last digit is 0, so the last digit of the number will be 0. Note that more than one number can satisfy the conditions for part a).

• March 14 at 1:37 pm

Question C: any multiple of 4

• March 14 at 1:37 pm

Option A: 3816547290
Option B: 314
Option C: 3.14

May I please win. Tomorrow is my birthday!

• March 14 at 1:37 pm

A. 3,816,547,290

C. Any multiple of 3 that is not divisible by 2.

• March 14 at 1:38 pm

Problem 1: 3816547290

• March 14 at 1:39 pm

• March 14 at 1:39 pm

A. Pie 3.14 is undef. It can be 3816547290 (10 integer) but n is not listed that is divisible or it can be other based on n so answer would be 3.14.
From reading glimpses of finite book you wrote,
B. 314
C. 3.14

• March 14 at 1:40 pm

3816547290 for problem A

• March 14 at 1:42 pm

Option B 314

• March 14 at 1:43 pm

Clearly I started too late… but it was a nice lunchtime challenge nonetheless.

• March 14 at 1:45 pm

The answer to option A = 1472589630

• March 14 at 1:45 pm

Option A: 3816547290

• March 14 at 1:47 pm

Any integer divisible by 3. Grouped in sets of 3

• March 14 at 1:47 pm

Option A: 3816547290 is the answer.

• March 14 at 1:48 pm

Option a: 3816547290

• March 14 at 1:48 pm

option a) 9843156720

• March 14 at 1:49 pm

Option A: 1234567890

• March 14 at 1:49 pm

Problem A: 3816547290

• March 14 at 1:49 pm

Question A: 3816547290
Question B: 314
Question C: 3.14

• March 14 at 1:50 pm

Problem A: 3816547290

• March 14 at 1:52 pm

Option A

9876543210

• March 14 at 1:53 pm

Option C: Any multiple of 6.

• March 14 at 1:56 pm

Answer A: 3816547290. I LOVE PIZZA HUT! 🙂 Happy Pi Day!

• March 14 at 1:56 pm

A: 3816547290

B: 314

C: 3.1415926535897923

• March 14 at 1:57 pm

• March 14 at 1:58 pm

The answer to Option A question is: 9876543210.

• March 14 at 2:00 pm

1472583690

• March 14 at 2:00 pm

Went ahead and brute forced A with a quick Android app out of bored. Super inefficient, but why not?

public class MainActivity extends AppCompatActivity {

int buckets[] = new int[10];

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

//I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?
@Override
public void run() {
for( long i = 1000000000L; i < 10000000000L; i++ ) {
Log.e( "Test", "testing: " + i);
if( validateUniqueDigits(i) ) {
if( checkNConsistency(i) ) {
Log.e("Pizza", "Number: " + i );
break;
}
}
}
}
}).run();
}

private boolean checkNConsistency(long num) {
long numFormedFromNDigits = 0;
for( int i = 0; i < 9; i++ ) {
numFormedFromNDigits += (long) ( getDigit(num,9-i) * (Math.pow(10, (9-i))));
if( numFormedFromNDigits / (Math.pow(10, (9-i))) % ((long) (i + 1)) != 0 ) {
return false;
}
}

return true;
}

private void initBuckets() {
for( int i = 0; i < 10; i++ ) {
buckets[i] = 0;
}
}

private boolean validateUniqueDigits(long num) {
initBuckets();
long tmp = 0;
for( int i = 0; i 1 )
return false;

num /= 10;
}

return true;
}

public int getDigit(long number, int digit) {
for( int i = 0; i < digit; i++ ) {
number /= 10;
}

return (int) number % 10;
}
}

• March 14 at 2:02 pm

Problem A: 3816547290
Problem C: 36

• March 14 at 2:02 pm

1472583690 – Option A

• March 14 at 2:03 pm

Option A:
1896543270
9816547230
7896543210
1836547290
9876543210
3816547290

• March 14 at 2:04 pm

Option C: 8

• March 14 at 2:04 pm

7613459280 for Problem A

• March 14 at 2:06 pm

3,816,547,290

• March 14 at 2:08 pm

YO MAMA!!!

• March 14 at 2:08 pm

Option A: 3816547290

• March 14 at 2:08 pm

Option A: 3816547290

• March 14 at 2:09 pm

Problem A: 1,796,543,280

• March 14 at 2:10 pm

Option A: 3816547290

• March 14 at 2:11 pm

option A – 3816547290

• March 14 at 2:11 pm

Option A: 0123456789 and ‘n’=3

• March 14 at 2:13 pm

option c- 9

• March 14 at 2:13 pm

Option b: 0

• March 14 at 2:15 pm

Option A: 1234567890
Option C: 18

• March 14 at 2:15 pm

option A – 3816547290

• March 14 at 2:16 pm

Option A:
3816547290

• March 14 at 2:18 pm

Question b : 240

• March 14 at 2:20 pm

option A: 3816547290

• March 14 at 2:20 pm

Option A: 3816547290

• March 14 at 2:21 pm

Question A: 9184567230

• March 14 at 2:21 pm

A)3816547290
B)314
C)3.14159265… (Pi)

• March 14 at 2:22 pm

For Option 2:
1+2+3=1X2X3=6
Therefore room 6

• March 14 at 2:23 pm

• March 14 at 2:23 pm

OptionB : room 314
Option C: 3.14

• March 14 at 2:26 pm

The answer to B is 314.
The answer to C is 31.4

• March 14 at 2:26 pm

option A : 3816547290

• March 14 at 2:27 pm

Option A=3816547290

• March 14 at 2:27 pm

Option: 3816547290
Option: 314
Option: 3.1415926535897932……….

• March 14 at 2:28 pm

option a: 7891234560

• March 14 at 2:28 pm

Option A: 3816547290

• March 14 at 2:29 pm

Option A: 4998

• March 14 at 2:29 pm

Option B

• March 14 at 2:30 pm

3816547290 for Problem A

• March 14 at 2:32 pm

Option A: 3816547290

• March 14 at 2:32 pm

A: 3816547290
B: 314
C: Pi

• March 14 at 2:32 pm

A: 3816547290

• March 14 at 2:33 pm

Option A: 3816547290

• March 14 at 2:33 pm

The answer for option A is 0145926387

• March 14 at 2:33 pm

Option A: 3816547290
Since only even numbers are divisible by even numbers, all even digits are even numbers
All odd digits are odd numbers.

Since only numbers divisible by 10 end in 0, d0=0. Only numbers ending in 5 or 0 are divisible by 5, so d5=5.

d1 + d2 + d3 = 0mod3

All 2-digit numbers divisible by 4 with an odd 10s end in 2 or 6, so d4=2,6.

For all 3 digit numbers divisible by 8 with pattern even-odd-even with no repeating digits end in 2 or 6, so d8=2,6. From this, I also know if d7=1,9 -> d8=6 and d7=3,7 -> d8=2.

The remaining even numbers are 4 and 8 for d2 and d6.

• March 14 at 2:35 pm

Question 1

3816547290

• March 14 at 2:36 pm

• March 14 at 2:36 pm

Option A answer is 9873516240 because…
n = 1: 0 is divisible by 1,
n = 2: 40 is divisible by 2,
n = 3: 240 is divisible by 3,
n = 4: 6240 is divisible by 4,
n = 5: 16240 is divisible by 5,
n = 6: 516240 is divisible by 6,
n = 7: 3516240 is divisible by 7,
n = 8: 73516240 is divisible by 8,
n = 9: 873516240 is divisible by 9,
n = 10: 9873516240 is divisible by 9

• March 14 at 2:36 pm

Option A: 3816547290 – Thanks!

• March 14 at 2:38 pm

A) 1022456789
Each digit divided by one ia itself.

B) Either not enough info or 314

C) 6, 12, 18, 24, 30

• March 14 at 2:38 pm

Q: C

• March 14 at 2:38 pm

Option a is 9345678120

• March 14 at 2:38 pm

Problem A : 3816547290
Problem B : 314
Problem C : 3.14

• March 14 at 2:39 pm

• March 14 at 2:39 pm

Option a: 1234567891
Option b: 100
Option c: 5 or7

• March 14 at 2:39 pm

Option A: 1472583690

• March 14 at 2:39 pm

For Option C:
The smallest number would be
7, including a central ring with 2 strands of 3. The series of numbers that comprise correct answers would then increase in multiples of 3 as additional strands of 3 are added.
So: 7,10,13,16,19…etc

• March 14 at 2:40 pm

I think I am reading this problem wrong. To me it sounds as if each number (1-10) needs to be divisible by the first integer of this 10 digit number. Only 1 would fit that bill. The order of the rest of the numbers shouldn’t matter as long as each digit is only used one time.

Answer B: Anything with a 0. 301 Perhaps.

If the room number had a zero in it, the product of the numbers would be zero no matter how many numbers there were. The zero would throw off the count.

• March 14 at 2:40 pm

For option A:

1,384,527,690

• March 14 at 2:42 pm

Problem A is 3816547290

• March 14 at 2:43 pm

Problem A: 3816547290

• March 14 at 2:43 pm

Option C: Half the numbers in the series {(3xn)/2} from n =3 to pi/2a, a= thickness of the key ring start number is based on the set up of the problem, you could go lower depending on how your read it. Since as the ring approaches max number it will become a sphere, a cross section shows that the max number of rings can be pi/2a where a=thickness of the key ring.

• March 14 at 2:44 pm

Option A – 3,816,547,290

• March 14 at 2:44 pm

OPTION A: 1234567890

• March 14 at 2:45 pm

Option A, could be taken kind of literally, and the answer can be Any series of numbers 1234567890. Why?
The question states: “I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?”
As long as n does not equal zero (and it states from 1 to 10), then the number will always be divisible by n. It never states that it needs to be divisible evenly. Also, it doesn’t specify if the number formed from the first n is starting from the left or the right. Most assume left to right, but is that actually the expected response. If it is from left to right and evenly: 3816547290, if it is from right to left and evenly: 9654873120

• March 14 at 2:46 pm

Option B = Room number 101

• March 14 at 2:46 pm

Problem A – 3816547290

• March 14 at 2:47 pm

Option A: 3816547290

• March 14 at 2:47 pm

• March 14 at 2:47 pm

Option B: 1
Option C: 6

• March 14 at 2:50 pm

Option B: 210

• March 14 at 2:50 pm

B ; you meet in the library

• March 14 at 2:50 pm

Option A: 3,816,547,290

Pizza would be used for student events @ Manual Arts High School 🙂

• March 14 at 2:53 pm

Option A: 1237894560

• March 14 at 2:53 pm

• March 14 at 2:54 pm

Option A: 9468315720

• March 14 at 2:54 pm

A: 3816547290

• March 14 at 2:54 pm

OptionA —3816547290

• March 14 at 2:54 pm

First question: 4918235760
Second 6
Third: 5

• March 14 at 2:54 pm

I believe Option A is: 3816547290

• March 14 at 2:55 pm

option A
0123456789

• March 14 at 2:56 pm

Problem A : 3816547290
Problem B : 314
Problem C : 3.141592

• March 14 at 2:58 pm

Option A: 3962840571

• March 14 at 2:58 pm

A. 3816547290
B 314
C 3.1415926535897932384626433832795028841971693993751058209949445923

• March 14 at 3:01 pm

9837645120 for option A

• March 14 at 3:02 pm

option a is 3816547290

• March 14 at 3:02 pm

Question b: One

• March 14 at 3:06 pm

Option A: 3816547290

• March 14 at 3:07 pm

Question a
9345678120

• March 14 at 3:07 pm

Why there are so many of the same answers for option A? There are many answers to this problem so either people are copying pasting the responses of others or are using same logic in their algorithms.

Here is one answer that I have not seen before for Option A:
9 2 1 6 5 4 3 8 7 0

• March 14 at 3:07 pm

Option B the answer is 1

• March 14 at 3:09 pm

The Answer to Option A = 0123456789

• March 14 at 3:12 pm

Option B 314
High School Student

• March 14 at 3:12 pm

Q.A= 3816547290

• March 14 at 3:12 pm

Option a- 3816547290
Option b- 12
Option c – 3.14 (pi)

• March 14 at 3:13 pm

3816547290 is B 😉

• March 14 at 3:14 pm

A) 3816547290
B) 314
C) 3.14

(One of these answers is not like the other)

• March 14 at 3:15 pm

• March 14 at 3:16 pm

Option A: 3.141592653

• March 14 at 3:17 pm

Option C: 3 to infinite

• March 14 at 3:17 pm

Option A: 3816547209

• March 14 at 3:18 pm

Question A: 3816547290

Question B: 314

Question C: 3.14

• March 14 at 3:18 pm

Option A is
3816547290

• March 14 at 3:20 pm

Option a is 3816547290

• March 14 at 3:21 pm

Option A: 3816547290

• March 14 at 3:22 pm

Option A: 3816547290
Option B: 314

• March 14 at 3:23 pm

Option A = 0123456789

• March 14 at 3:24 pm

Option B is room 3.14

• March 14 at 3:27 pm

B) Room #502

• March 14 at 3:27 pm

Option C is 12

• March 14 at 3:30 pm

14

• March 14 at 3:30 pm

Option a is
1234567890

• March 14 at 3:31 pm

Option A
3816547290

• March 14 at 3:33 pm

Option A: 3816547290
Option B: 315
Option C: 17

• March 14 at 3:35 pm

Option A is 3816547290

• March 14 at 3:35 pm

Dear Dr. Conway, thank you for the fun. Here is my proposed solution to Option C:

There may be 3 or 9 rings. You specified circular rings, but if we allow for a small amount of eccentricity (and given the identity of the owner of the key-rings, I think a small amount of eccentricity should be expected!), we may construct from three (slightly eccentric elliptical) rings a set of Borromean rings. We may end there, but the wording of the problem says that some triple may be distinguishable from other triples, which suggests more than one triple (although the single triple is vacuously distinguishable from the other non-existent triples, so three rings works as a solution).

Therefore from two more sets of three rings each, construct two new sets of Borromean rings, and link these new sets such as to form a “super set” of Borromean rings, with each “super ring” composed of a set of Borromean rings.

The result is nine key rings, no two of which are actually linked (so any pair of two is indistinguishable from any other pair), but for which there are some triples which are distinguishable from other triples: namely, there are three sets of Borromean rings, each a triple which is distinguishable from any triple which is not a set of Borromean rings.

Finally, whereas this method could number-theoretically be extended for more rings, such as by creating a super-super-set and using a total of 27 rings—or in general 3^n for any positive integer n—in actuality, it would be physically impossible for more than 9 individual metal key rings of about 2″ diameter.

Hence, I claim that you have 3 or, more likely, 9 key rings in your jumble.

peace,
Sky Waterpeace

As a post-script, I suggest you examine all the associated keys and identify which you currently use on a regular basis, since when I found myself with a set of 9 jumbled key-rings in the form of a multiple Borromean rings, it turns out that almost all of the keys were to unidentifiable in their utility. I removed all keys with imaginary uses (those for which I only imagined that I knew what locks they went with), and I somehow ended up with e^(i * pi) keys remaining—that is, I had misplaced the one key I actually needed and so had -1 many which were identifiable.

• March 14 at 3:35 pm

The answer to option B is: Room 314

• March 14 at 3:36 pm

optionA: 3816547290

• March 14 at 3:36 pm

Question A: 3816547290
*As n goes to 10, the division of n+1 is proven*

Question B: 314
*lol Also the same room as my Calc II class XD*

Question C: 3.14
*Pretty straight forward, can be proven with visual aid and basic division.

• March 14 at 3:40 pm

Option B = 6
1+2+3 = 6
1 *2*3=6

• March 14 at 3:40 pm

The answer to C is 6

• March 14 at 3:41 pm

A) is 1
B) is 1
C) is 3.141592653587993

• March 14 at 3:41 pm

Option C: The possible numbers are infinite (Undefined).
Example, never ending chain of key rings.

• March 14 at 3:41 pm

Option B the answer is 314

High School Student

• March 14 at 3:43 pm

Option A ; 3816547290

Option B : 314

Option C : 5

• March 14 at 3:44 pm

Question A: 3816547290
*As n goes to 10, the division of n+1 is proven*

Question B: 314
*lol Also the same room as my Calc II class XD*

Question C: 3.14
*Pretty straight forward, can be proven with visual aid and basic division.

• March 14 at 3:44 pm

Another Answer to A is 1,472,583,690.
1, the 1st digit, divided by 1 =1.
14, the first 2 digits, divided by 2 =7.
147, the first 3 digits, divided by 3 =49.
1472, the first 4 digits, divided by 4=368.
14725, the first 5 digits, divided by 5 =2945.
147258, the first 6 digits, can be divided by 6 =24543.
1472583, the first 7 digits, can be divided by 7 =210369.
14725836, the first 8 digits, can be divided by 8 =1840730.
147258369, the first 9 digits, can be divided by 9 =16362041.
1472583690, all 10 digits, can be divided by 10 =147258369.
I’m surprised no one came up with this one and all the others are the 3 trillion number, I feel google maybe have influenced these.

• March 14 at 3:44 pm

Answer for Option A is 3816547290

• March 14 at 3:45 pm

Option A is 3816547290
Option B is 314
Option C is 3.141592653589793238462643383279502884197169399375105820974944592307.
Now let’s see that pizza 🙂

• March 14 at 3:45 pm

option b: “not enough information”

• March 14 at 3:47 pm

1234759680

• March 14 at 3:47 pm

Option A;
3816547290

• March 14 at 3:48 pm

question b: 314

• March 14 at 3:50 pm

C – 11

• March 14 at 3:51 pm

Option A: 3816547290

Option C: 12

• March 14 at 3:54 pm

Option A

• March 14 at 3:54 pm

Option B : 1

• March 14 at 3:58 pm

A: 3816457290
C: 4

• March 14 at 3:59 pm

• March 14 at 4:00 pm

For those who care about Option A. Definitive proof that there is only 1 answer and it is: 3816547290. This simple java program checks every possibility.

public class PiDay
{

public static void main(String[] args)
{
ArrayList matches = new ArrayList();
for (int i = 0; i < 10; i++)
{
int[] match = new int[10];
for (int j = 0; j < 10; j++)
match[j] = 0;
match[0] = i;
}
for (int i = 2; i <= 10; i++)
{
ArrayList newMatches = new ArrayList();
for (int[] match : matches)
{
long nbr = getNbr(match, i);
for (int j = 0; j <= 9; j++)
{
if ((nbr + j) % i == 0 && !used(match, j, i))
{
int[] newMatch = new int[10];
for (int k = 0; k < 10; k++)
newMatch[k] = match[k];
newMatch[i – 1] = j;
}
}
}
matches = newMatches;
newMatches = new ArrayList();
}
System.out.println("Solution:");
for (int[] match : matches)
{
for (int x : match)
System.out.print(x);
System.out.println();
}
}

public static long getNbr(int[] match, int pos)
{
String nbr = "";
for (int i = 0; i < pos; i++)
nbr += match[i];
return Long.parseLong(nbr);
}

public static boolean used(int[] count, int nbr, int pos)
{
for (int i = 0; i < pos – 1; i++)
{
if (count[i] == nbr)
return true;
}
return false;
}

}

• March 14 at 4:03 pm

4 problem C

• March 14 at 4:04 pm

Option C:
6n, where n is any positive integer

• March 14 at 4:05 pm

A. 3816547290
B. 12
C. 3.14

• March 14 at 4:06 pm

Problem C:
Infinite

• March 14 at 4:09 pm

• March 14 at 4:09 pm

Option A is 1239567840.
It is a multiple of 1 because every number is.
It’s a multiple of 2, 4, and 8 because the last two digits form a multiple of 2, 4, and 8.
It is a multiple of 3 and 9 because the sum of the digits is a multiple of 3 and 9.
It is a multiple of 5 and 10 because it ends with 0.
It is a multiple of 7 because 177,081,120 × 7 = 1239567840. Hooray for proof by guess and check.

• March 14 at 4:10 pm

Option A: 3816547290

• March 14 at 4:11 pm

C is 3*n, where n is a positive integer.
One possibility for A is 3627548190.
A has a few solutions

• March 14 at 4:12 pm

The answer to C is 13

• March 14 at 4:15 pm

Option c. 31

• March 14 at 4:15 pm

Problem C: 3^n
where n is the number of triple of keyrings.

• March 14 at 4:18 pm

• March 14 at 4:19 pm

First one, last one has to be 0, middle has to be 5, and numbers alternate odd and even… a very simple case by case analysis will show only possible answer is 3816547290

• March 14 at 4:19 pm

Problem C: 3^n
where n is the number of triples in the keyring.

• March 14 at 4:22 pm

C)6, 12, 18, 24, 30….both divisible by 2 and 3 which would give you pairs and triples

• March 14 at 4:23 pm

Option A has two answers: 9,876,543,210 and 1,896,543,270.

— Doron Holzer

• March 14 at 4:24 pm

Option C: 9 or more

• March 14 at 4:25 pm

3816547290

• March 14 at 4:25 pm

Option C: Infinite possible options.

• March 14 at 4:26 pm

the answer to option A is 3816547290.

• March 14 at 4:26 pm

Answer to option b is 314

• March 14 at 4:28 pm

C)136
Because 1 set of rings 1 triple and 1 double

• March 14 at 4:32 pm

Option C

(3.14(2)) x 3 = 18.84

18 total rings

• March 14 at 4:34 pm

• March 14 at 4:34 pm

Option C: 6 or 10

• March 14 at 4:34 pm

Option A = 3816547290

• March 14 at 4:35 pm

Problem C: 3 since you can go either from the left or the right.

• March 14 at 4:36 pm

3816547290

Room 13 product 36 (6*6*1 = 2*2*9)

I can distinguish some triples from other triples in a chain of links with any even number of rings > 3

• March 14 at 4:38 pm

Okay, for someone who wants the proof.

Option A:

The last digit must be zero, for the tenth to be divisible by 10. *********0
For the fifth to be divisible by 5, that must be 5 since 0 is used ****5****0
For the fourth digit to be divisible by 4, the 3/4th must be also.
12 16 24 28 32 36 48 64 68 72 76 84 92 96
The second digit must be even: 2,4,6,8.
For the 3rd to be divisible by 3, 1st-2nd-3rd must be also.

But that makes the 4-5th digits either 25, 45, 65, 85. Because the 6th digit
must make a number divisible by 3 (sum of digits divisible by 3), then digits
4-5-6 must also be divisible by 3, and even also. So if the 4th digit is
2, the 6th must be 8; if the 4th digit is 4, the 6th must be 6; if the
4th digit is 6, the 6th must be 4; if the 4th digit is 8, the 6th is 2.

So that only leaves
123654 – which has no 7th digit
129654 – with a 7th digit of 7
147258 – with a 7th digit of 3
183654 – nothing
189654 – nothing
321654 – nothing
327654 – nothing
369258 – with a 7th digit of 4
381654 – with a 7th digit of 7
387654 – nothing
723654 – nothing
729654 – with a 7th digit of 1
741258 – nothing
783654 – with a 7th digit of 2
789654

For the eighth digit to be divisible by 8, the 6-7-8th must be also.

The 6th is 4 or 8, so the 7th-8th must be divisible by 4.
12, 16, 24, 28, 32, 36, 64, 68, 72, 76, 84, 92, 96
However, if the 6th is 4, then 4 and 6 are both eliminated;
if the 6th is 8, both 2 and 8 are eliminated, resulting in a 6th-7th-8th of:

412, 428, 432, 472, 492 816, 836, 864, 876, 896

However, 412, 428, 492, 836, and 876 still aren’t divisible by 8, so..

12965472 fails (duplicate 2)
14725836 fails because 836 is not divisible by 8
3692584 fails (no digit is divisible by 8)
38165472
72965412 fails (duplicate 2)
78365428 fails (duplicate 8)

That leaves only one digit left, a 9:

3816547290

• March 14 at 4:39 pm

A. 3816547290
B. 314
C. 19

• March 14 at 4:41 pm

Option C: Any multiple of 6.

• March 14 at 4:41 pm

A: 3816547290

• March 14 at 4:42 pm

Solving Problem C:
9 key rings. 3 rings together makes a triangle. Connecting 2 rings to each ‘point’ of the triangle creates a triple ring section. up/down & left/right are indistinguishable.

• March 14 at 4:42 pm

Option c= y = sin θ = 2 tan ½θ / ( 1 + tan2½θ ) and x = cos θ = ( 1 – tan2 ½θ ) / ( 1 + tan2½θ )

• March 14 at 4:44 pm

Option C: 12

• March 14 at 4:49 pm

Question A: 1834567290

• March 14 at 4:49 pm

The answer to Option A: 9876543210

• March 14 at 4:50 pm

Option C: 18
3*3*2 is the only combination of numbers I can come up with from the problem text

• March 14 at 4:50 pm

Problem C is 9

• March 14 at 4:50 pm

For problem C, I’m going to assume that you are forced to distinguish between two triples on the basis of an up-down arrangement type.
In that case, you need at least seven keys (think of a hexagonal arrangement of keychains with six circles representing vertices, and a circle representing center. However, this is not possible, as the chains must be intertwined, and in this arrangement, they are merely touching boundaries.

• March 14 at 4:51 pm

Option A
7836549210

• March 14 at 4:52 pm

C:9

• March 14 at 4:52 pm

Option C: 54 3^3 *2

• March 14 at 4:54 pm

Option c: 17

• March 14 at 4:54 pm

Option A:
3816547290

• March 14 at 4:54 pm

6 to question three

• March 14 at 4:55 pm

OPtion C: At least 7

• March 14 at 4:56 pm

Option C: infinite

• March 14 at 4:57 pm

Option B: 0
Option C: 81

• March 14 at 4:59 pm

Problem C: Any number divisible by 5

• March 14 at 5:00 pm

48 possible for problem C

• March 14 at 5:00 pm

Option C: Any even number 6 or greater!

• March 14 at 5:00 pm

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

(3/1 = 1) (38/2 = 19) (381/3=127) (3816/4 = 954) (38165/5 = 7633) (381654/6 = 63609) (3816547/7 = 545221) (38165472/8 = 4770684) (381654729/9 = 42406081) (3816547290/10 = 381654729)

Option A: 3816547290

—————————————————————————————————————————————————————————————-
Our school’s puzzle-club meets in one of the schoolrooms every Friday after school.

Last Friday, one of the members said, “I’ve hidden a list of numbers in this envelope that add up to the number of this room.” A girl said, “That’s obviously not enough information to determine the number of the room. If you told us the number of numbers in the envelope and their product, would that be enough to work them all out?”

He (after scribbling for some time): “No.” She (after scribbling for some more time): “well, at least I’ve worked out their product.”

What is the number of the school room we meet in?”

Option B: (Assuming she has worked out the product of the room, pie is a product, and pi is a homophone, and seeing there cant be a room that is on the 3.14th floor, that would be impossible unless you’re at hogwarts. The closest whole number to pi is 3, but 3 isn’t a product of a whole number. assuming the first 3 numbers of pi which is 314, the three would equal 8 if added (1+3+4 = 8), but would equal 12 if multiplied (1*3*4 = 12) and seeing we’re also talking about a product of a number which the girl had worked out, the only product of 3*1*4 is 12.)
—————————————————————————————————————————————————————————————-
My key-rings are metal circles of diameter about two inches. They are all linked together in a strange jumble, so that try as I might, I can’t tell any pair from any other pair.

However, I can tell some triple from other triples, even though I’ve never been able to distinguish left from right. What are the possible numbers of key-rings in this jumble?

(Any number factorable by two and three which are the first two prime numbers. could equal 6, 12,18,24,32 and so on until infinity, but a 2 inch key chain cant hold infinite loops, rather there cant be infinite materials on this world for production. Since the diameter is 2, the radius is 1, which makes the circumference 6.28 (Tau). An average 2″ thick key ring is .125″ thick. In one ring, on the inner circle of the ring 50 rings can be held without an overlap. the closest 2*3 factor would be 48 since 48/3=16 and 48/2=24. Therefore there are 48 key chains on this key chain.)

• March 14 at 5:01 pm

I Don’t Know Where Else To Comment This So , Option C 3,6,9,12

• March 14 at 5:01 pm

Option C: Can’t tell pairs apart, meaning more than one pair – so at least four. Can tell triple from other triples, so at least three triples, bringing the number up to 13. After that, you can continue to add 2 or 3 onto that number… 15, 16, 17, 18 which continues on forever. So 13 or more.

• March 14 at 5:02 pm

Option c: 12

• March 14 at 5:02 pm

Option B: 314

• March 14 at 5:04 pm

A.PI
B.PI
C.PI

• March 14 at 5:04 pm

Option A: 3816547290

• March 14 at 5:04 pm

Option B is 6 right?

• March 14 at 5:05 pm

Option C: 4

• March 14 at 5:06 pm

3

• March 14 at 5:08 pm

• March 14 at 5:08 pm

A hotel with an infinite number of rooms was filled with an infinite number of reservations, and there is no vacancy, and all the guests check in. Without having guests double up or being inconvenienced after check in, how is the clerk able to find a room for more guests as needed. Show work for full credit(can be verbal).

• March 14 at 5:09 pm

The answer to option C is 2(n+2) where n is a natural number than makes this equation end in a palindrome. For example n=49 which leads us to 2*51 = 101. This is because he cannot tell left from right.

• March 14 at 5:09 pm

I Don’t Know Where Else To Comment This So , Option C Has To Be At Least ,6 , And Option A Is 3816547290 🙂 🙂

• March 14 at 5:09 pm

Option C :13

• March 14 at 5:09 pm

Option b: 24

• March 14 at 5:09 pm

option c: 24 or 42

• March 14 at 5:11 pm

Option C:6

• March 14 at 5:13 pm

C: any prime number greater than 2

• March 14 at 5:13 pm

The answer to option C is any number 6 or greater.

• March 14 at 5:14 pm

The answer to option C is 2(n+2) where n leads us to a palindrome. For example n=49.

• March 14 at 5:19 pm

Option B: Room 8

• March 14 at 5:19 pm

Option c: 20

• March 14 at 5:20 pm

A:1

• March 14 at 5:20 pm

C: all prime numbers greater than 2

• March 14 at 5:22 pm

Option C: Every number divisible by 3 greater than 3.

• March 14 at 5:22 pm

• March 14 at 5:22 pm

A: 3816547290

• March 14 at 5:22 pm

• March 14 at 5:22 pm

• March 14 at 5:24 pm

a:0123456789
b: 0
c: 3

• March 14 at 5:26 pm

Option C : 33

• March 14 at 5:27 pm

Option C: five combinations

12
121
112
212
221

• March 14 at 5:28 pm

C: either 7 or 13 rings

• March 14 at 5:28 pm

1,023,456,789

• March 14 at 5:29 pm

Option C : 5,6 and 7 rings

• March 14 at 5:29 pm

Option C : 36 ways

• March 14 at 5:29 pm

Option A: 1,023,456,789

• March 14 at 5:30 pm

It has to be greater than 6. That’s 4 for the 2 pairs and 3 for a lone triple, and the rest are all singletons. If there is 1 triple, it can always be distinguished.

• March 14 at 5:30 pm

Option C: every multiple of 6 greater than 12

• March 14 at 5:30 pm

Option C : 36 ways.

• March 14 at 5:31 pm

Option A: 3816547290

• March 14 at 5:33 pm

Question C: You have 15 rings.

• March 14 at 5:33 pm

Option C: 10 is the first value and all successive values are 10n where n is a positive integer.

• March 14 at 5:34 pm

3.1415926536

• March 14 at 5:35 pm

Option C:
The answer is a series. 4n where n is the number of triples, when n>1. Therefore the answer is 8, 12, 16, 20, 24, etc…. to infinity.

• March 14 at 5:35 pm

Question A: 9876351240

• March 14 at 5:35 pm

c is 19

• March 14 at 5:36 pm

Option A.)
3816547290

• March 14 at 5:36 pm

Answer c: Triangular numbers such that the sequence continues of all possibilities that are in finite countable dispositions to represent the finite amount of key chains. The numbers of key chains possible are 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, etc.

• March 14 at 5:37 pm

6 for c

• March 14 at 5:39 pm

• March 14 at 5:39 pm

A. is 3816547290
B. Is 314
C. Is 4 or more rings

• March 14 at 5:39 pm

• March 14 at 5:41 pm

OPTION C 3 IS THE ANSWER

• March 14 at 5:43 pm

option c 6,12,18,24

• March 14 at 5:46 pm

Option A is 3816547290 because each group of numbers has to be divisible by a number in the sequence

• March 14 at 5:46 pm

C:15, +5, etc.

• March 14 at 5:47 pm

question c: 9 rings

• March 14 at 5:47 pm

There have to be at least two pairs – four, and at least three triples – nine, which adds up to thirteen. After that, one would only need to add 2 or 3 to the number to add a pair or a triple. So….15, 16, 17, 18, 19, etc. There can be 13 rings, or fifteen or more rings on an integer.

• March 14 at 5:49 pm

Option A-987654312
Option B-314
Option C-6

• March 14 at 5:49 pm

Option c: 19 or 13

• March 14 at 5:49 pm

Option C: Multiples of 3. The rings are linked in 3’s in this formation https://upload.wikimedia.org/wikipedia/commons/0/07/Borromean-rings_minimal-overlap.svg

• March 14 at 5:50 pm

Option A: 9123567480

• March 14 at 5:50 pm

I am answering question C. My answer is every odd number greater than or equal to 3.

• March 14 at 5:51 pm

Option c is 3.1415926535897932…

• March 14 at 5:51 pm

Option C: The answer is 15

• March 14 at 5:52 pm

3816547290 #PiDay My birthday!!!

• March 14 at 5:54 pm

Problem C: 18 key-rings

• March 14 at 5:54 pm

Problem C: 8 rings

• March 14 at 5:55 pm

A 123
B 1234567890
C 10

• March 14 at 5:57 pm

3816547290 for problem A

• March 14 at 5:57 pm

Problem A 3816547290

• March 14 at 5:57 pm

Option A
3816547290

• March 14 at 5:57 pm

Option A is 3816547290

• March 14 at 5:57 pm

C is 26 there are two pairs that triple that have tripples so its 2+3+3+3+3=26. And a is 3816547290

• March 14 at 5:58 pm

Option C: 66, 222, 252, 282, 414, 444, 474, 606, 666, 696, 828, 858, 888

• March 14 at 5:58 pm

B: 117

• March 14 at 5:59 pm

The answer to Question C is any integer above the number 3 could be a possible answer to this question. So starting with 4 then 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21… Infinity

• March 14 at 5:59 pm

Option C.
The answer must be an even number that is divisible 3. So the answer is 6.

• March 14 at 6:00 pm

Option c is 6.

• March 14 at 6:00 pm

C= 6

• March 14 at 6:01 pm

Answer to question A is 3816547290

• March 14 at 6:02 pm

I think the answer to problem C is :you have to have at least 15

• March 14 at 6:05 pm

• March 14 at 6:05 pm

Option C: Any integer that is 4 or higher.

• March 14 at 6:08 pm

Problem A: 3816547290

• March 14 at 6:08 pm

Option c 6

• March 14 at 6:08 pm

For option C: In order to have two pairs that one cannot tell apart there must be at least three rings. These rings would be in a line, A-B-C, so that one could not tell the AB pair from the BC pair. To be able to have at least two triplets which one can tell apart, there must be four rings. In this case a fourth ring would be connected to both B and C. The ABC triplet would be distinguished from the BCD triplet, as for the latter, each ring is linked to two other rings, however in the first triplet rings A and C are each connected to only one ring in the pattern. One can add rings without changing these two properties. Therefor the set of key rings has four or more key rings in it.

• March 14 at 6:09 pm

Option C:
Any multiple of 6, 12 or higher.

• March 14 at 6:13 pm

C- any number; not enough information to solve the problem

• March 14 at 6:13 pm

Option C: 3, 9, 15, 27, 39, etc.

• March 14 at 6:14 pm

Option C: Any number with an odd number of digits (with a minimum of 3 digits) with a repeating pattern of two numbers, Ex., 1212121. It is the same left to right as it is right to left, every two you count in either direction would be the same (the pairs), and every three would be different from the one next to it.

• March 14 at 6:15 pm

The possible number of key rings is any number greater that is 6 or greater
because he clearly says that he can tell “some triples from other triples” indicating that there is more that one triple, and two triples is 6. the It can be any number more than six because he never mentions that all the rings fit into a triple and that he only can tell some triples from others and also the wording seems as though the triples don’t overlap

• March 14 at 6:19 pm

• March 14 at 6:21 pm

Problem C: There is an infinite number of possibilities. The first part, the key rings being 2 diameters has no relevance in this question. Just because you can tell some triples from others, and not pairs, has no effect on determining the amount of possible key rings. Because it implies that there are indeed pairs, and you can only tell triples apart, you can not accurately determine the amount of pairs. Therefore, you could have an infinite amount of pairs, and not know it because you can’t tell them apart.

• March 14 at 6:23 pm

Option c the answer is 5 but configured as 4 rings in a T and one ring joining 3 others. The ring with 3 connected rings is then unique.

• March 14 at 6:24 pm

Option C: Any number with an odd number of digits (minimum three digits) with a repeating pattern of two numbers, ex. 1212121. It is the same left to right as it is right to left, every two you count in either direction would be the same (the pairs), and every three would be different from the one next to it.

• March 14 at 6:25 pm

option c: 4

• March 14 at 6:28 pm

C Any odd number 5 or greater

• March 14 at 6:28 pm

C=9

• March 14 at 6:29 pm

• March 14 at 6:31 pm

n is greater than or equal to 3

• March 14 at 6:31 pm

B is 314

• March 14 at 6:31 pm

Option C
8
There are 8 rings because when 2 pi triple bonds connect there are 8 rings

• March 14 at 6:32 pm

// Solution A:
// By Kirk Weedman: KD7IRS: http://www.hdlexpress.com

#include
#include
#include

void clear_used_digits(void);
int is_divisible(long, long);
void next_digit(int);
void show_unused_digits(void);
void show_solution(void);

int used_digit[10];
int digit_value[10];
int number;

main()
{

int n;

clear_used_digits(); // clear from 1 – 10
//last digit is 0 in order for 10 digit number to be divisible by 10
//fifth digit is 5 in order for 5 digit number to be divisible by 5

for (n = 1; n <= 9; n++) // 9 digit number
{
printf("Starting off with %d\n",n);
next_digit(n);
}

printf ("Number = %d", number);
}

void next_digit(int n) // which digit we're on
{
int k, j;
long number;
int found;

number = 0;
for (k = 1; k < n; k++) // calculate current value
{
number *= 10;
number += digit_value[k];
}
number *= 10;

found = 0;
for (j = 1; (j “, number+j);
used_digit[j] = 1;
digit_value[n] = j;
if (n == 9)
{
show_solution();
exit(-1); // just abort the program
}
next_digit(n+1);
used_digit[j] = 0;
found = 0;
}
}
}
if (!found && (n < 10))
{
show_unused_digits();
}
}
int is_divisible(long val, long divide) // ie. 125
{
long r;

r = val/divide;

return ((r*divide == val) ? 1 : 0);
}

void clear_used_digits(void)
{
int k;

for (k = 1; k <= 9; k++)
used_digit[k] = 0;
}

void show_unused_digits(void)
{
int k;

for (k = 1; k <= 9; k++)
if (used_digit[k] == 0)
printf("[%d]", k);
}

void show_solution()
{
int k;

printf("\n Solution: ");
for (k = 1; k <= 9; k++)
printf("%d", digit_value[k]);
printf("0\n"); // last digit has to be 0 so only have to solve other 9
}

• March 14 at 6:32 pm

In the number 3816547290 the first digit would be 0, not 3. Therefore 290 is not divisible by 3. The correct answer should be 9678345120.

• March 14 at 6:32 pm

Option C: 10

• March 14 at 6:32 pm

The answer to question A is .) 1472583690

• March 14 at 6:32 pm

The answer to question c is 33. You can tell some jumbled from the others. 11 triples or just 3 and 3.But you can’t tell the difference between left and right cause they are both the same number.

• March 14 at 6:33 pm

Problem C: 18 Key—rings

• March 14 at 6:35 pm

Option A: 1836547290

• March 14 at 6:38 pm

Option C:
since the triples are distinguishable and the pairs are not, a possible jumble would be a cylindrical arrangement made by linking n sets of six linked rings (3×2) together such that the cylinder-like jumble is three rings high and 2n rings in circumference where n >= 2.
thus, the possible numbers of rings in the jumble are all real numbers equal to 6n where n is an integer greater than or equal to 2.

• March 14 at 6:39 pm

Option c: 4

• March 14 at 6:39 pm

Problem C: Any multiple of 2, Any multiple of 3, OR Any Multiple of 5. So, 2, 4, 6 8..etc etc. or 3 6 9 12, etc etc, or 5 10 15 20 etc etc
I read it wrong at first.

• March 14 at 6:40 pm

Option c is 66. It is palidromaic and it is divisible by both 2 and 3.

• March 14 at 6:40 pm

My guess for question 3 is 14.

• March 14 at 6:40 pm

Any odd number greater than 5

• March 14 at 6:41 pm

Any integer greater than 4 that is not a multiple of 2 nor a multiple of 3.

• March 14 at 6:41 pm

C os 6n-3
But i dont know

• March 14 at 6:41 pm

Option C: 2(n+1) + 3(n+1) where n=even positive number. 15, 25, 35, etc

• March 14 at 6:42 pm

Option C: 6

• March 14 at 6:42 pm

Problem A: 3816547290

• March 14 at 6:43 pm

A:0123456789
B:48
C:9

• March 14 at 6:43 pm

The answer to question number3 is 56.52

• March 14 at 6:45 pm

Option c: Any multiple of 6

6, 12, 18, 24, ect.

Can be paired and tripled

• March 14 at 6:45 pm

C. Any number greater than or equal to 9

• March 14 at 6:45 pm

C:4

• March 14 at 6:45 pm

Option C:
There must be at least 3 sets of 3 (identify some triple from other triples) rings. Since they are all connected and you cannot identify pairs, that implies that every ring is at least connected to one other ring (a set of two). Hence, your jumble of rings could be any number that is greater than 10 that is divisible by both 2 and 3 (12, 18, 24…) or in other words, any multiple of 6 that is greater than 10.

• March 14 at 6:46 pm

Option c is 66. It is palidromaic and can be divided by both 2 and 3.

• March 14 at 6:46 pm

C is 6n-3

• March 14 at 6:46 pm

Option C: 845

• March 14 at 6:46 pm

Option C : the answer is 12

• March 14 at 6:47 pm

The answer to C is 6 or another multiple of 3 or 2… So 6, 12, 18

• March 14 at 6:49 pm

Problem A 3816547290

• March 14 at 6:52 pm

option C 3.14

• March 14 at 6:52 pm

Option C: 32

• March 14 at 6:53 pm

The answer to letter c is 4 or more rings. I’m assuming since you’re using key rings, you could bend them in a manner that would allow you to create Borromean rings. Technically, with Borromean rings, no two rings are connected, but you can see a triple. Adding a fourth ring would allow you to see more than one triple. If you would like to see a depiction, please email me. Thanks

• March 14 at 6:59 pm

Answer to C:There are 3 key rings

• March 14 at 7:01 pm

Option C: We know there are pairs and triples, so the answer has to be divisible by 2 and 3. The fewest possible rings would be 12. The way the question is worded, there are at least 3 triples, but to be divisible by two, there would have to be at least 4 sets of triples (12 total rings). Other potential solutions would include 18, 24, 30, 36, and so on increasing by 6 each time. This will result in answers that are divisible by 2 and 3.

• March 14 at 7:03 pm

Part C is 35 rings (or 12 triple ring groups)

• March 14 at 7:04 pm

question c is 6

• March 14 at 7:05 pm

Option C: The number can be any that can be divided by 2 and 3. For example 6, 12, 18, 24, and so on.

• March 14 at 7:06 pm

To have indistinguishable pairs and distinguishable triples, each pair must consist of two rings, one with x total links and one with y total links. The triples will therefore either have x/y/x or y/x/y groupings, while the pairs will all be x/y. The smallest number of rings that can be used is 7, and each integer is a possibility higher than that as adding the different number of total connections to complete the circle, x+y, is the total number of rings needed. 2 and 3 can form a circle of interconnecting ring satisfying the pairs, but no distinct triples can be discerned as there are not enough rings for two sets. 2 and 4 allow enough rings, but both triples are 2/4/2 in construction. 3 and 4 allow for separate triples of 3/4/3 and 4/3/4 to be isolated while satisfying the pairs as well.

• March 14 at 7:08 pm

Option c

5 and every odd number greater than 5 would work

• March 14 at 7:09 pm

Problem A: 4
Problem C: 9

• March 14 at 7:09 pm

Option C: 3,4,5

• March 14 at 7:10 pm

c-9

• March 14 at 7:11 pm

A 3816547290

• March 14 at 7:11 pm

Option C: If it takes 55.0 Joules (J) to raise the temperature of a 9.00 grams (g) piece of unknown metal from 13.0 degrees Celsius (°C) to 24.9 degrees Celsius (°C), then the specific heat of the unknown metal is 0.513539 Joules per gram • degree Celsius (J/g•°C). The given data on the problem are the amount of heat energy gained or lost by the unknown metal, Q = 55.0 Joules (J), mass of the unknown metal, m = 9.00 grams (g), and the change in temperature, ΔT = 24.9 degrees Celsius (°C) – 13.0 degrees Celsius (°C) = 11.9 degrees Celsius (°C). With these given, you can now solve the problem using the formula Q = mcΔT, where c is the specific heat of the unknown metal. Using the formula above, you can rearrange the equation to get tehe value of c. The equation will now become c = Q/mΔT. Substituting the given to this equation will result to c = 55.0 Joules (J) / 9.00 grams (g) • 11.9 degrees Celsius (°C). The result will be c = 0.513539 Joules per gram • degree Celsius (J/g•°C). Therefore, the specific heat of the unknown metal is 0.513539 Joules per gram • degree Celsius (J/g•°C). The closest metal with specific heat of 0.513539 Joules per gram • degree Celsius (J/g•°C) is diamond, thus the unknown metal would be diamond. Diamond in its solid state, whose chemical formula is C, has a specific heat of 0.519 Joules per gram • degree Celsius (J/g•°C). Another closest metal would be Titanium. Its solid state has a specific heat of 0.523 Joules per gram • degree Celsius (J/g•°C).

• March 14 at 7:13 pm

Option C: 3, 4, 5

• March 14 at 7:18 pm

Option C — 18 — It has to be divisible by both 2 and 3 because I cannot distinguish between one pair (2) and another pair (2) but I can tell between one triple (3) and other triples (at least 6 more or (3+6=9 at a minimum). Since 9 is not divisible by 2, the answer is 12 since it is the lowest number both divisible by 2 and 3. The next possible answer would be 18 followed by 24, etc.

• March 14 at 7:22 pm

Option C: There is no number satisfying these properties.

• March 14 at 7:25 pm

Option C: 6, 12, 18, 24…..

• March 14 at 7:25 pm

B=0

• March 14 at 7:26 pm

A- 3816547290
B- 4
C- any whole number greater than or equal to 4

• March 14 at 7:26 pm

c:5, because if you draw a picture and you start with three rings you can only add two more rings to the middle of the original ring or the groups wouldn’t be distinguishable.

• March 14 at 7:26 pm

Question C:

Any even number greater than 4.

• March 14 at 7:28 pm

Question C:

Any number greater than 4.

• March 14 at 7:30 pm

c:5

• March 14 at 7:31 pm

C: 4

• March 14 at 7:33 pm

3816547290 because 3 is divisible by 1, 38 is divisible by 2, 381 is divisible by 3, 3816 is divisible by 4, 38165 is divisible, 381654 is divisible by 6, 3816547 is divisible by 7, 38165472 is divisible by 8, 381654729 is divisible by 9, and finally 3816547290 is divisible by 10

• March 14 at 7:35 pm

Answer for Problem/Option C: Infinite Amount.

• March 14 at 7:38 pm

A 3816547290
B 314
C 6

• March 14 at 7:42 pm

Option c is 9
3 distinct 3’s but no distinct 2’s because it’s not an even number.

• March 14 at 7:42 pm

Option A. The answer is 3816547290.

• March 14 at 7:43 pm

Options C is 69 rings. No joke.

• March 14 at 7:45 pm

question #1 is 3
question #2 is 1
question#3 is 4

• March 14 at 7:46 pm

question 3/C

• March 14 at 7:46 pm

option C: 12,15,18,21 etc

• March 14 at 7:47 pm

Answer C: Any whole number between 0 and infinity.

• March 14 at 7:48 pm

OPTION A: 3816547290 because 3 is divisible by 1, 38 is divisible by 2, 381 is divisible by 3, 3816 is divisible by 4, 38165 is divisible by 5, 381654 is divisible by 6, 3816547 is divisible by 7, 38165472 is divisible by 8, 381654729 is divisible by 9, and finally 3816547290 is divisible by 10

• March 14 at 7:52 pm

Option A: 3816547290
Option B: 0(there can be any numbers in the list- example:1*2*0=0)
Option C: 6

• March 14 at 7:52 pm

• March 14 at 7:53 pm

Any number >6 Ex:

00
000
00

Has 2 indistinguishable chains of 2 and 3 distinct chains of 3. At this point we can add to the protruding chain on the right making it a chain of 4, giving us 2 indistinguishable chains of 2 and 2 distinct chains of 3, and a chain of 4 (cheeky, but not prohibited) anything 6 or less makes producing multiple distinct chains of 3 impossible.

I look forward to three years of free pizza! Cheers!

• March 14 at 7:54 pm

Option C: Multiple of 6. 6, 12, 18, 24, etc.

• March 14 at 7:55 pm

Option C: 3 + 6x

• March 14 at 7:55 pm

Question C: 9=<9+(6*|n|)
In interval notation I think it's [9, 9+(6*n)) I don't remember how to do this lmao.

• March 14 at 7:56 pm

For question A everyone seems to have started from the left and went right. In attempting this problem I decided to go from the right, and start at the ones place, and go left. As this made much more sense to me.
Question A then has two answers:
9,315,648,720
9,135,648,720.

• March 14 at 7:58 pm

Option A:3816547290
Option B:0
Option C:6

• March 14 at 7:58 pm

C: You could have 6 or 12 rings.

The jumble must be a 3D structure. I use polyhedral geometry. It could be either an octahedron or icosahedron. Each ring corresponds to a vertex, and each connection between rings corresponds to an edge. The faces of these shapes are equilateral triangles, which would correspond to one type of triple (3 rings connected to each other, think trinity symbol). There would also be triples of rings that are only connected to the middle ring. Since the entire shape has repeating symmetry, the double rings are never unique and there is no “left” or “right”. Sidenote: If it is were not a requirement that there be two distinct types of triples, you could also have a total of 3 rings arranged in a triangle.

• March 14 at 7:59 pm

C: You could have 6 or 12 rings.

The jumble must be a 3D structure. I use polyhedral geometry. It could be either an octahedron or icosahedron. Each ring corresponds to a vertex, and each connection between rings corresponds to an edge. The faces of these shapes are equilateral triangles, which would correspond to one type of triple (3 rings connected to each other, think trinity symbol). There would also be triples of rings that are only connected to the middle ring. Since the entire shape has repeating symmetry, the double rings are never unique and there is no “left” or “right”. Sidenote: If it were not a requirement that there be two distinct types of triples, you could also have a total of 3 rings arranged in a triangle.

• March 14 at 8:00 pm

Problem c : 7

• March 14 at 8:02 pm

Option A: 1234567890

• March 14 at 8:02 pm

Option C is 18

• March 14 at 8:02 pm

Any odd number 5 or greater.

• March 14 at 8:03 pm

Option C: Only 5 () () ()
() ()

• March 14 at 8:04 pm

Option C 8

• March 14 at 8:04 pm

Question C: Any odd number 5 or greater.

• March 14 at 8:06 pm

A. 3816547290

• March 14 at 8:08 pm

Option C: max 15 rings
There are 6 distinct triples comprised of two linked and one non-linked ring.
Where three are linked, there are two triples- the chain and the trifoil.
However, the non linked rings must be connected to something so they are removed from the count.
The rest depends upon spacing and how many triplets one ring is involved in.

• March 14 at 8:09 pm

A: 3816547290
B: 314
C: 3.14

• March 14 at 8:11 pm

4 arranged as a pyramid

• March 14 at 8:12 pm

• March 14 at 8:12 pm

The question has individual clauses that give us information as to the parameters that we must use to define our answer. If we look at each clause, we can reap it for the information that it provides and use that to make a list we can use for our answer.

“They are all linked together in a strange jumble,”

All the rings are connected. There are not two groups of rings.

“I can’t tell any pair from any other pair.”

Any two connected rings must be indistinguishable from any other two rings.

“I can tell some triple from other triples,”

Any group of three connected rings and any other group of three connected rings must be distinguishable at least once in the system.

“I’ve never been able to distinguish left from right.”

The orientation of the rings doesn’t matter. Three rings vertical=three rings turned on edge diagonal.

Now that we know what the system parameters are, we can develop a ring system that meets all of these with the smallest amount possible.

If we arrange two rings interlocked, and then add a third so that it is connected to both, finally adding one more that attaches to only one ring, we get a system that fulfills the requirements. Select any pair, and it is indistinguishable from any other. Select the three rings that are all connected, and that triplet is one where all rings are connected from every other ring. This is different from the triplet that is formed when you select two center rings, and the one outer one. Here we receive a chain of three rings, different from the triplet that we first formed. The triplet distinction is not based on orientation. The rings are all linked.

This system uses four rings.

We can have a system with n rings, where n>=4 and natural, if we add the ring on to the fourth ring that we added to the center triplet, and any more to this growing chain. All previous clauses remain intact.

The exact nature of our answer, though is “What are the possible numbers of key-rings in this jumble?” The possible numbers are then all natural numbers greater than or equal to 4.

• March 14 at 8:12 pm

The answer to a is 3816547290

• March 14 at 8:13 pm

Any number, 12 or greater, which is evenly divisible by both 2 and 3.

• March 14 at 8:14 pm

So i got 3816547290 for problem A
Then for problem B i got 314
and lastly for problem i got 3.141592 i dont know if we were supposed to round but theres my answer hope i when thanks

• March 14 at 8:15 pm

My answer to C: any number whose factors are 3 alone; that is 3^n where n is an integer greater than 1.

• March 14 at 8:15 pm

Option C: After 3 and 6, any even number can be made to form that jumble

• March 14 at 8:16 pm

So i got 3816547290 for problem A
Then for problem B i got 314
and lastly for problem C
i got 3.141592 i dont know if we were supposed to round but theres my answer hope i when thanks

• March 14 at 8:18 pm

Option B: 12

• March 14 at 8:20 pm

option c:15 rings

• March 14 at 8:20 pm

Option C: Any number, 12 or greater, which is evenly divisible by both 2 and 3.

• March 14 at 8:20 pm

Question A: 3816547290

Hope everyone worked this out rather than copy it! 10th digit must be 0. 5th must be 0 or 5 (therefor 5 because 0 is used). Digits 2,4,6,8, and 10 must be even, leaving only odd for rest. Using this info and that first 3 digits add to be divisible by 3, can find 20 permutations for first 3. 30 permutations remain for first 4. 5th is same due to being 5. 19 valid permutations of 6 digits. 7 possible of 7 digits and only 1 possible with 8 digits. 9th is a gimme because digits all add up to 45 which is divisible by 9 and thus all possible combinations not using 0 (in 10th spot) are valid!

• March 14 at 8:21 pm

Option b 101

• March 14 at 8:22 pm

Problem A:38165472

• March 14 at 8:25 pm

Since we need to have both pairs and triples it must be divisible by 2 & 3. However, we need to have ‘triple’ distinguished from other ‘triples’ soooo we must have at least 3 triples. 3 triples is not divisible by 2. So, the possible solutions are any even multiples of 3 beginning at 12 ….(and of course… there should be a maximum reasonable answer to at least be able to ‘distinguish’ among pairs and triples. Although any number that is an even multiple of 3 beginning at 12 works, I would cap around 108 ?!?)

• March 14 at 8:27 pm

Option c: Any integer, n>= 5, will allow more than one triple to occur without sharing a double.

• March 14 at 8:28 pm

Option C: Infinite number

• March 14 at 8:29 pm

option C: multiples of 12

• March 14 at 8:30 pm

Option C: unlimited, but answers start with 6 and then you add 3 for the next potential integer (6, 9, 12, 15, 18, 21, 24, 27 ect.)

• March 14 at 8:30 pm

Option a:
9876543210

• March 14 at 8:30 pm

• March 14 at 8:31 pm

Option C: 7 rings

• March 14 at 8:31 pm

• March 14 at 8:32 pm

• March 14 at 8:33 pm

A. 3816547290
3 is an integer of 1
38 is an integer of 2
381 is an integer of 3
3816 is an integer of 4
38165 is an integer of 5
381654 is an integer of 6
3816547 is an integer of 7
38165472 is an integer of 8
381654729 is an integer of 9
AND
3816547290 is an integer of 10
B. Is most likely 3.14
and
C. is any number just not negatives

• March 14 at 8:33 pm

The answer to Option C is any multiple of 6 starting at 12

• March 14 at 8:33 pm

2in keyring, 52 of them linked together with 3 triples able to show and 1 linked together on the sides.

• March 14 at 8:36 pm

• March 14 at 8:36 pm

option c 5,9,13, 17, 21, 25 etc… if triples and doubles are keys on the rings

• March 14 at 8:39 pm

Any prime number for question C

• March 14 at 8:41 pm

Option C: Any multiple of 10

• March 14 at 8:45 pm

The number of rings is of the form: S(n)=S(n-1) + 3 Where S(n) is an integer from 6 to infinity. My answer is, in other words, the number of rings could be 6,9,12,15…. n .

• March 14 at 8:45 pm

Option C : 2

• March 14 at 8:45 pm

Option A is 3816547290.

• March 14 at 8:47 pm

12,14,28

• March 14 at 8:48 pm

Option C: Infinity!

• March 14 at 8:50 pm

The answer is 13 for c

• March 14 at 8:51 pm

The answer to option C is 11 rings

• March 14 at 8:52 pm

• March 14 at 8:52 pm

For problem B, you’re in room 4

• March 14 at 8:54 pm

It must be a whole number, and negative numbers don’t mean anything, so we’re in the space of positive integers.

The problem describes a “jumble,” so we must assume that it is at least two rings. As zero means no rings and one is “a ring,” not a jumble.

Depending on how you want to read the problem, 2 is a possibility. Given the same pair over and over, he’d never be able to tell the difference, so it could be a “jumble” to him.

3 wouldn’t work because there’s only one tuple he’d see….but, it says that he can only tell “some” from others. If we accept that it’s possible that he can’t determine that he’s seen the same tuple, he’d never know if he was seeing different tuples, or if he just recognized it sometimes.

The problem makes no mention of four or more, but those are just multiple arrangements of tuples.

I conclude that the subject would be confused on any number of rings two or greater.

Therefore, to answer the question “What are the possible numbers of key-rings in this jumble?” there are infinite possibilities (2, 3, 4, etc.).

Even if you assume there must be at least three rings and he’d always recognize a tuple, then the smallest acceptable jumble would be 4: he’d see the same four tuples over and over again, but he would’t know if he was seeing the only tuples, or if he was just having bad luck.

• March 14 at 8:55 pm

Total # of rings = (# of triples)*3 – [((# of triples)*1) -1]

1 triple = 3 rings
2 triples = 5 rings
3 triples = 7 rings
4 triples = 9 rings
…………

• March 14 at 8:56 pm

Option c: from 4 to an infinite amount of rings.

• March 14 at 8:56 pm

Problem A:
3816547290

• March 14 at 8:58 pm

option c: 444, 636, 696, 828, 888…any multiple of 12 that reads the same backward and forward

• March 14 at 8:58 pm

• March 14 at 9:00 pm

For option C: I think it’s possibly 12, 18, 24, 30, or 36. All of these numbers can be pairs, triples, and still seems…. reasonable.

• March 14 at 9:00 pm

Option A: 3816547290.

• March 14 at 9:01 pm

c=minimum of 4 to infinity

• March 14 at 9:03 pm

Option C: Multiples of 12 that read the same forward and backward (444, 636, 696, 828, 888 … )

• March 14 at 9:11 pm

Option A: 3816547290
Option B: 314
Option C: 36

• March 14 at 9:12 pm

3.1416

• March 14 at 9:16 pm

• March 14 at 9:16 pm

Option C: f(x)=3^3n where n is any positive whole integer and n cannot = 0

• March 14 at 9:17 pm

question c = 10, 12 or higher. the question implies at least two pairs and at least two triples. 2×2=4, 3×2=6, 6+4=10. 11 is not an option because the question says they are all hooked meaning there are no singles. From here, you can add 2s or 3s indefinitely.

• March 14 at 9:17 pm

Question C is 6.

• March 14 at 9:17 pm

C=3.1416

• March 14 at 9:19 pm

C is 8 because there are 8 slices in a pizza

• March 14 at 9:21 pm

Question C: 12,15,18,21,24…

• March 14 at 9:24 pm

Option A: 1,234,759,680

• March 14 at 9:26 pm

• March 14 at 9:27 pm

12 key rings for problem c

• March 14 at 9:27 pm

OPTION C: 2 key rings

• March 14 at 9:29 pm

Option A: 9,876,351,240

• March 14 at 9:30 pm

Option C: 36

• March 14 at 9:33 pm

Problem A

3816547290

• March 14 at 9:34 pm

Option c: anywhere from 0 through infinity.

• March 14 at 9:36 pm

Option C: 18

• March 14 at 9:37 pm

Question C: 4n, 4n+n,4n+n+n,4n+n+n+n…

Where “n” is any number 3 or greater. Start with 4n then keep adding an “n” .

• March 14 at 9:39 pm

Option A :3816547290

• March 14 at 9:40 pm

The answer to C is 35.

• March 14 at 9:42 pm

Option C: Infinity

• March 14 at 9:43 pm

Option C’s answer: all multiples of 6

• March 14 at 9:43 pm

Option C: any multiple of 3 that is odd (9, 15, etc.)

• March 14 at 9:43 pm

Option C: Infinity

• March 14 at 9:44 pm

Option c is 5 or more
With one ring in the middle with at least 4 rings attached only to the middle ring and not to each other.
No pairs could be separated because they would all have the middle ring in common. But pulling one from the top and one from the bottom would form a triple with the center ring. And you could pull one from each side at the same time to form another triple. The question states that some triple from other triples. Not that he can tell tripleS from others triples. So using that logic the triples could overlap. But really what do I know. lol Thought I’d give it a shot.

• March 14 at 9:50 pm

option c: 12

• March 14 at 9:50 pm

Option c:
3x + (x-1)
Where x is the amount of triples and x cannot be less than 2

• March 14 at 9:51 pm

A: 3816547290

• March 14 at 9:51 pm

Option C: 3, 12…

• March 14 at 9:53 pm

Problem A:
Six solutions here…
3816547290
7412589630
7896543210
9816543270
9816547230
9876543210

Solved using divisibility rules, permutation principles, and a bit of paper and pencil. Nice to see that four of the six solutions I found were posted by less than two dozen people!

• March 14 at 9:54 pm

OPTION: C

6. 5 links linked and surrounding a center link, while being linked to the link to its left and right. No pair of links can be found different from the other, and it is possible to tell the difference between three links right to left from a triangle of links.

• March 14 at 9:54 pm

For option C:
The number of key rings would need to be divisible by both 2 & 3 in order to pair them by pairs and triples. This would include numbers such as 6,12,18, etc. However, the answer would be infinite because numbers are infinite. The number of options would go on forever.

• March 14 at 9:54 pm

The answer to C is 35

• March 14 at 9:57 pm

Option C:

9,12,15,18, continuous multiples of 3. There are at least 3 rings. Diameter seems irrelevant.

• March 14 at 9:58 pm

OPTION C: 2 KEY-RINGS

• March 14 at 9:59 pm

Option A: 3816547290

• March 14 at 10:00 pm

Option C.
Answer is 6… a pair of triplets.

• March 14 at 10:03 pm

Option C: 3.00

• March 14 at 10:04 pm

OPTION C IS 35

• March 14 at 10:04 pm

C: For C, think of a single ring in the middle called A, then attach five rings to that ring A, called B, C, D, E and F. Now to each B and C, attach another ring, we will make these lowercase, so b and c. You now have A, B, C ,D, E, F, b and c, 8 rings. You have triples in ADE, ADF, AEF, ABb, ACc. So this is true for 8 rings. Some triples ABb and ACc are distinguishable from the others. This is good, now it we add another ring to A, call it G, the math still works, ABb and ACc are still different from the rest ( ADE, ADF, AEF, AGF, …). This can continue since we are only adding rings to A. So the answer is simply, any number of rings greater than 8, meaning the number of combinations is infinite.

• March 14 at 10:06 pm

3 or 5

• March 14 at 10:08 pm

Option C

The answer is any multiple of 6 because there are pairs and triples.

• March 14 at 10:16 pm

option c:
The answer to option C is that there is 4 or MORE chain rings

The reason so is because we know there must be at the minimum 2 triples
Considering that the triples over lap (because the author cannot distinguish between left and right) and then that must mean that the lowest number of rings to create 2 triples is 4.

• March 14 at 10:16 pm

Option C: 12

• March 14 at 10:20 pm

option b, 1

• March 14 at 10:21 pm

Option C) 3.1416

• March 14 at 10:22 pm

Option C – 11

• March 14 at 10:22 pm

Problem A: 3816547290

• March 14 at 10:23 pm

Option C: 7 Keyrings

• March 14 at 10:29 pm

Option C:10

• March 14 at 10:30 pm

C: 13 keys

• March 14 at 10:30 pm

The answer to the question “A” is 3816547290 because it is the only number that makes sense.

• March 14 at 10:31 pm

C: 13 rings

• March 14 at 10:32 pm

Option C: Number of keyrings is any integer greater than or equal to 9 (n >= 9)

• March 14 at 10:36 pm

option c: 7 is less than or equal to x is less than infinity

• March 14 at 10:36 pm

The reason so is because we know there must be at the minimum 2 triples
Considering that the triples over lap (because the author cannot distinguish between left and right) and then that must mean that the lowest number of rings to create 2 triples is 4.

• March 14 at 10:38 pm

C: 18

• March 14 at 10:40 pm

The reason so is because we know there must be at the minimum 2 triples
Considering that the triples over lap (because the author cannot distinguish between left and right) and then that must mean that the lowest number of rings to create 2 triples is 4.

• March 14 at 10:42 pm

Y=(x/2)^2
where x is >=6 and x=2K where k is any integer. Y=total amount of rings. X=total number of counted triples .
X will always be even since he is double counting the triples since he cant distinguish left from right. Since the rings are 2in in diameter then the figures made by the rings will be a square. Thus y=(x/2)^2.

• March 14 at 10:44 pm

Option C = 18^n where n is an integer 1 or greater

• March 14 at 10:45 pm

The answer to Option C is 6

• March 14 at 10:47 pm

Answer to C: any number that is divisible by 6

• March 14 at 10:48 pm

C: any number that is divisible by 6

• March 14 at 10:48 pm

Let N be a prime number such that N, N^N+1, N^N+2,…N^Nth * any multiple of 3 is a possible solution.

• March 14 at 10:50 pm

C: 6, 12, 18, etc.

• March 14 at 10:51 pm

Option C is greater than 9

• March 14 at 10:52 pm

C: (3^x) + (6^x) where x > or equal to 1. only natural numbers.

• March 14 at 10:52 pm

Option C: multiples of 6 rings: 2 sets of 3 rings in a triangle with parallel connections, 4 sets of 3 rings in a pyramid, then solids with identical parallel faces with 3 rings per vertex. Each ring is in a triple and a double and three rings are in a triangle or line.

• March 14 at 10:53 pm

problem A=3816547290

• March 14 at 10:55 pm

Option C: from 3 to infinity. (Like how Archimedes approximated the value pi by using circumscribed and inscribed polygons then dividing those polygons into equal area triangles, we use the sides of the polygon as point where a ring could exist. Doubling the number of polygon sides each time approaching infinity.) The number of points within a 3.14 in. sq. circle where two circles are tangent can approach infinity without more data as we don’t know how much area is taken up by that point of the 2 tangent circles. We know that at least 3 must exist as a given triplet is observable but past that there is not a limit with the given data.

• March 14 at 10:55 pm

Option 1: The number 2,736,951,840 is divisible by 1,2,3,4,5,6,7,8,9 and 10.

• March 14 at 11:36 pm

Oops, I meant 7,126,951,840
since 0/1=0,
40/2=20,
840/3=280,
1840/4=460,
51840/5=10368,
951840/6=158640,
6951840/7=993120,
26951840/8=3368980,
126951840/9=14105760, and
7126951840/10=712695184
Happy Pi Day!

• March 14 at 10:56 pm

Greater than or equal to 5

• March 14 at 10:56 pm

Personally, I got Option A as 3816547290
I solved it within 2 hours.
Option B, I got 10
Option C, I got all multiples of six.

• March 14 at 10:56 pm

The answer to Question A is 3816547290

• March 14 at 10:59 pm

For question C = 413

• March 14 at 10:59 pm

C: 0 key ring, no combination exists where you cannot tell your left and right.

• March 14 at 10:59 pm

A=3816547290

• March 14 at 10:59 pm

Option c greater than or equal to 5

• March 14 at 11:00 pm

Option C:

9, 12, 15, 18 , continuous multiples of 3.There are at least 3 ring sets of 3. Diameter appears to irrelevant.

• March 14 at 11:01 pm

7836541290 for A

• March 14 at 11:02 pm

Problem B: 1

• March 14 at 11:02 pm

Option C retry: at least 6 rings, even number sets of 3 in a triangle connected linearly.

• March 14 at 11:03 pm

Option C: The answer is 4. There are 3 rings connected to each 1 ring. Therefore, no matter how you look at the rings you can tell there are three rings.

• March 14 at 11:04 pm

Option C: 6 + 4n, where n is an integer greater than 0.

Because you can’t tell any two rings apart, that means each connection must be symmetric. Likewise, there can’t be any dangling rings. As there are triples, that means some pairs of connections are distinguishable. The obvious indistinguishable forms a large loop. Two have different kinds of triples, that loop needs to have nodes extending out from it. But to not tell any two rings apart, that means they all have to have the same number of connections, which I determined to be 3. Starting from the smallest loop (a triangle), adding another node connecting to each does not work. All triples are indistinguishable. Symmetry must be maintained, so this involves adding one node on to each triangle point. It cannot be double linked, or you’ll fall into the problem of triples being indistinguishable again. Each of those three nodes are connectable (so think triangle inside of triangle, with the points connected. From here, you must basically divide connections in half, but the new rings can’t intersect with any of the previous rings. Adding one ring leaves a dangling connection. Inserting two rings and having them connected fulfills having 3 connections. However, the overall symmetry is broken and you can probably distinguish pairs. Inserting three rings however, means they all have one dangling connection which can be grabbed by a fourth.

As before, all the rings are connected symmetrically, so any pair cannot be distinguished from one another. Triples may be distinguished as either forming a line or all being connected. So, I guess that makes my answer (6 + 4n). where n is an integer greater than 0. Found diagrams for 6 and 10.

• March 14 at 11:05 pm

For Question C: 3

• March 14 at 11:06 pm

3816547290 for problem A

• March 14 at 11:07 pm

Any pair from any other pair. At least 2+2. Any Triple from any other Triple”s”. 3+3+3+(3n)+2n. Therefore 2+2+3+3+3+n. C=any number≥13

• March 14 at 11:09 pm

Answer C is 4: One Center Ring surrounded by and attached individually to 3 separate rings

• March 14 at 11:12 pm

Option C retry: at least 6 rings, even number sets of 3 in a triangle connected linearly, or vertices of a solid with an even number of points.

• March 14 at 11:15 pm

Option A: The number is 3816547290
Option B: 0
Option C: Prime numbers greater than 13
Pretty cool challenge, definitely enjoyed it even if I don’t win

• March 14 at 11:17 pm

Option c: there are five key-rings, because i think the keys are represented as the olympic logo.

• March 14 at 11:18 pm

3 is divisible by 1
38 is divisible by 2
381 is divisible by 3
3816 i divisible by 4
38165 is divisible by 5
381654 is divisible by 6
3816547 is divisible by 7
38165472 is divisible by 8
381654729 is divisible by 9
3816547290 is divisible by 10
So the answer to option A is 3816547290

• March 14 at 11:19 pm

Solving C:

I believe the answer to be at at least 6 rings, one positioned at the center and five interlocking adjacently around. Only three rings are involved in each intersection, (so triples are able to be partitioned by lifting the rings at individual intersections, but not pairs). The shape created by this formation is a circle, making distinguishing left from right nearly impossible. (If only 5 rings are used, this nulls the circle effect). The max circumference of the circle formed with the rings is 18.84 inches, ( with a diameter of 6 inches using three rings spread completely out). This makes the max amount of rings jumbled up to be 10, (one inside and nine arranged around). Any more than this will hinder discerning triples. So my final answer is 6-10 rings.

• March 14 at 11:22 pm

Attempt at question c: