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*

3816547290 for problem A

option c: 5

Problem A: 3816547290

Answer C is 3.141592653589793238462643383279502884197169399375105820974944592307…….

Answer C: 27

Option a is 3816547290

Option b is 314

Option c is 3.1415926535897932……….

Option A: 38165472

Option A=3816547290

Option B=314

Option C=3.141592653589793238462643383279502884197169399375105820974944592307

Question A: 3816547290

Question B: 314

Question C: 3.14

First question answer is 5

Second question answer is 12

Third question answer is 3

Option A: 3816547290

3.141592653589793238462643383279502884197169399375105820974944592307…

3816547290 for option A

A: 3816547290

Answering Option A: 3816547290.

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

3816547290

The answer to C is 3 or more.

Option A:

3816547290

Wrong! Obviously the answer must be an integer value.

answer to A 1274569830

Answer C is any multiple of 3 plus any multiple of 4

Option B is any room with a 1 or 0 in it

Answer B is room 0

Answer To Question 3: Any multiple of six

Option C is 32

Problem C = 3

Answer for A is 3816547290

Answer to A is 3816547290

Answer to C: 6, 12, 18, 24, 30, 36…

The answer to problem C is 48.

Option C: 9

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.

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.

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

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.

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.

Option C: 114

Answer C is 4

Option B: 0

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

Answer A: 3816547290

Answer B: 314

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

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.

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.

Option A’s answer is: 3816547290

Option A: 3816547290

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

Answer C is 3.141592653589793238462643383279502884197169399375105820974944592307

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

answer to a is 3816547290

A. 75891361915

A: 1472583690 works too!!

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

Òption C: at least 7

Option A:

3816547290

Even though so many other people have predicted this number.

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.

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

How are you reaching that number specifically?

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

A: 3816547290

option B is 314

B: 632

C:7

Problem A.) 3816547290

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.

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.

Option C is 9.

A: 3816547290

B.314

C: 3.141592

Option A is 3816547290 or 4998

Problem C: 19

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

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

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.

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

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

Answer B = 0

room # 1

A:3816547290

B:Room 314

C:3.14……….

C: = 6561 combinations

Option A: 3816547290

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

Option a: 1234567890

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…

How to comment

Option a is 3816547290

3816547290 is the answer to question A

Question c is 3.141592

Option a: 3.14159

Option b: 27

Option c: 6

A: unique answer is 3816547290.

B: unique answer is 19.

C: still working on it… difficult

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.

Option A : 3816547290

Problem a: 3816547290

For option A– 3816547290

Option a: 1,234,759,680

Option A: 3816547290

Option A: 3816547290

1010101010 problem a

No. Problem b

3 problem c

C. 23

C: 26 rings

8

Answer for Option C is 8.

C: 1

Answer to C is 6

A:1

B

D

H

P

B

D

H

P

Problem C:15

Question C: 5 Rings

Problem C :15

Option c: any multiple of 6

Like 6, 12, 18, 24

Option c: any multiple of 6

6, 12, 18, 24, ect

Answer to C: 15

Option C is any perfect square except 4. (9,16,25,36…)

They are arranged in the shape of a square

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

Option C: 48

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.

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

69

Question C: An even number 6 or greater.

Option C: 440169

option C: any multiple of 5

7 and 22

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

Option C is 10

Option C is 11

Option C: 9

C is 54.

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.

Option C: 28

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)

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.

Option C: is 12

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

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.

Answer c. 27

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 .

Answer c is 11

Option A=3816547290

Option B=21

Option C= 5

Option A: is 3816547290

A: 3816547290

B. 16

Answer to B is 29

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.

optionC: 2

C:9

ANSWER TO “C” = 3

This doesn’t work for n = 8

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

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.

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

C. 3.14159265358979

Option a – 3816547290

b. 2

Option B is 4

B. 4

B: 100

Option A = 3816547290

A: 3816547290

Option C

Option C:

Any multiple of 6.

I.e. 6, 12, 18, …..

C

Any multiple of 6

A is 3816547290

Option B: is 6 that’s what I got

problem A 1896547290

Is not divisible by 7. 1896547/7= 270935.2857

Room 12 – Option B

Option B: Room #1.

A . 3816547290

B . 314

C . 3.141592653589

314

Option a: 3816547290

3.141592653589

3.141592653589

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

Question A: 3816547290

Question B: 314

Question C: 3.14

Option B: 314

Cheater

Get off my nutz

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:

N must start with 147.

If N starts with 14725:

N must start with 147258.

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:

N must start with 183654.

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 start with 381654.

N must start with 3816547.

N must start with 38165472.

N must start with 381654729.

N must be 3816547290.

Through casework, we can find that no other N works. Therefore, the answer is 3816547290.

holy shit m8

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

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!

Option A: 4998

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

woah watch your language

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

Option a. 7560

3816547290 option A

Option B – Room 6

22

a: 3816547290

Problem B: 22

38165444290 problem a

Option B…3

3816547290 problem a

Option A: 9786450213

Option B: 202

Option C: 1

9735846120 for problem a

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

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.

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.

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.

Option A – 9345678120

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.

Problem C: 64

Option A: 3816547290

B 314

Problem A: 3816547290

ANSWER for problem a is 5

A 1234567890

B Room 11

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

Answer to Option B: any multiple of 10

option 1: 3816549270 option 2 :0 option 3 6

A: 7293648150

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

Option A: 3816547290

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

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.

C 3

its 3.14

Problem A

2520000000

For problem B the room is 314

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.

Option A 3816547290

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

Option A

3816547290

Option B= 0

Option b =2

Option B – 101

Option A: 1

The answer to option A is 3816547290.

Option a: 7,831,549,620

Option A: 3,816,547,290

Option A 3816547290

Answer to C. is 3

Option a – 3816547290

option a is 10!

Option b: 314

OPTION B; 1

OPTION C: 12

option A

9731428560

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

Problem b: room 0

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

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

Option B the answer is 4

Option A: 3816547290

Option B: 2

Option C: multiples of 6

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.

I am not seeing that number divisible by 4 though.

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

option B: 0

b 13

Option A: 3816547290

Option B: Room 7

A. 3816547290

Option C : 3

C:3

For problem a

1827364590

A. 3816547290

Option b: 314

Option c: 31

option A 3816547290

Option A: 3816547290

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

Option c. 9 rings with 2in. Diameter

Option C is 1234567890

A) 3816547290

B)283

C)3.14159265358979

problem c answer 1

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

A=3816547290

C=6

Option A= 3816547290

9632581470

9632581470 for option A

Question A:3816547290

Such a number doea not exist.

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

Option B i believe to be is 36 rooms

Option C i also believe ro be is 6

option B 1

option C 27

C is 7

B i believe to be is 36 rooms

C i also believe ro be is 6

Option A: 3816547290

A is 3816547290

Option A

1234567890

B is 36

C is 6

1

B I beleve is 36

C I believe is 6

Option C- Multiples of 3

Option C 6

A) 1834567290

B) 10

C) multiples of 7

C: 12 +6n

Option A: The number is 1234567890

Option A: 3816547290

OPTION A: 9,876,543,210

3816547290 problem A(option A)

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

Option A: 3816547290

Option A – 9,876,351,240

1252566450

A 4

B 1

C 3

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

Option A

1032547698

N=2

Option A

1,111,111,111

Option A: 3816547290

1264564800 for Option A

Option A: 1234567890

Noah Taylor googled it:

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.

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.

Option A

3816547290

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

Well done.

Option A: 3,816,547,290

problem A: 3816547290

Answer b: 63

Option A: 9657834120

Option C: i think 9, 15, 21, 27, 33….

option a

321654987

Option C answer= 12

Option A=1

A: 3816547290

B: 10

C: multiples of 7

One answer to A is 3816547290.

For option A: 3816547290

Ooption A=1

3816547290 for A.

Option A…..1234567890

A. 1234759680 c. 12

Option A: 3816547290

option a 2468109573

Problem B : 6

option A

Option A=1

Option A = 3816547290

Option A is 3816547290

Option C: Multiples of 9

B.) 1

Option A = 3816547290

Option C:

Any multiple of 6.

I.e. 6, 12, 18, …..

Option C is 3, if the links are intertwined.

Option A: 3816547290

For option a the answer is 2520

Option a

3816547290

Option A: 120456789

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

The answer to problem A is 0123456789

answer to a is: 3816547290

Option The answer is 0123456789

0123456789 for option a

The answer to B=314

Option AThe answer is 0123456789

Option b: room 0

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

Option c: 33

Opton A 3816547290

Option B is 55

3816547290 for A

The answer to problem A is 1234567890

The answer to problem B is 6.

The answer to problem C is 5

option b : room zero

Option A is:

3816547290

Option B is:

7

Option C is:

27

Option A: 3816547290

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.

Option A: 5312948760

Option A is 314

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!

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.

Option C: 32

Option a is 3816547290

Option A answer is 1,234,568,160

Option A

1234759680

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.

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???

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.

Option A: 6234513789

Option A: 3816547290

Ooops, looks like no free pizza at our house.

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

A: 3816547290

A:3816547290

B: 4

C: 3

The answer for option A is 87687687.

Option B: 310

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

A is. 3816547290

Option C: 9

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

Classroom # 8

Option A is: 9,523,478,160

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

option B 102

Option A:

9, 356, 712, 480

the answer to A is 3816547290

Problem A

1023456789

Option C – 5 rings

A. 3816547290

B. Room 1

C. 5 Rings

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!

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.”

Option A = 9657834120

Option A: 1234567890

Option B 1 or room 1

1- 3816547290

2-314

3- 3.14159265…..

Option A – 3816547290

3816547290

A: 3816547290

Option c 3.141592

3816547920 option a

Option A: 987654321

Here is a small perl program.

Answer is 3816547290.

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);

wrong

Answer to B = 11

A.

3816547290

Option C: π itself.

Option A: 3816547290

Option A: 9735846120

Answer to question no. 1 is 1111111111

3816547290 for problem A

Option A:

1,4,9,16,25,36,49,64,81,100

One for option B

Option B: 12

Option c is 3 rings

0

B: room 1

1987654320

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

Option A: 3816547290

B. 314

C. 3.1415926535897932……… infinity

Option A is 3816547290.

Option A: 1

Option B: 1

Option C: 5

Answer to option A:

3816547290

Option B: 9

Option B is 120

3816547290

Option A=3816547290

Option B=314

Option C=3.14159265358979323846…..

Option C: 9

Option A: 3816547290

1111111111

3

8

Option B: 101

Option A is 3816547290

Problem A: 3,816,547,290

Option b is 100

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

For option A I got 9876543210

Option A answer is 3601598247.

Option C is 27.

3816547290 for question A

Option C

10

Option A = 1234567890

Option B = 0

Option C = 6

Option A=3816547290

Option B=314

Option C=3.141592653589793238462643383279502884197169399375105820974944592307

at least I think these are the right answers …

Option 2: 8

Option A: 3816547290

Option B = 45

A) 1,234,759,680

question #1 123487560

Option A: 3816547290

sorry….. 1234875609

Option a

3816547290

For Option/Problem A

3691287540

Option A: 3816547290

Option B:109

OptionC:15

Option A 3816547290

Option B: 14

Option b: 21

Option C: 72

Answer A is definitely 3816547290

Option A.) 3816547290

A: 3816547290

B: 4

C: 3.1415926535

Option B

4

Option A: 3816547290

Option c: any odd multiple of 3

(3,9,15,21, etc.)

A: 3816547290

Problem A: 3816547290

really oh my gosh iam in college that is so easy

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

Option A=3816547290

Option B=314

Option C=3.141592653589793238462643383279502884197169399375105820974944592307

Problem A 3816547290

The room number is 314

B: 4 or 5

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.

C is 1023456789

Option C: One

Option A: 3816547290

option A= 3816547290

option B= 314

option C= 11

C=9

3816547290 that is for problem A.

Option C: 6

Option A = 3816547290 Is my answer

3816547290

Problem A: 3816547290

A: 1,234,759,680

B: pi

C; 27

Option B. Answer is 0.

Option A to pie contest

2496543210

option c 9

Option B: The room number is 12

(Assuming the numbers in the list are all positive integers)

option a – 1357111317

option b – 1

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

Option A also has 5698743120 as a solution

Option A: 3816547290

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

A) 3816547290

Option B: room 0

Option A: 8,697,543,120

Option A is 8,697,543,120

Option B: 1

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.

For option A:

1234567890

A) 9657834120

3816547290 for option A

A=3816547290

B=3.14

C=3.14159265389

Option B: 314

Option C:3.14159265359

are you sure i don’t know

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

Option c

Pair = 2

Triple=3

L/r=2

2*3*2=12

All multiples of 12….12, 24, 36, 48…

option b answer 5

Option A: 3.1415926535

Option A answer: 3816547290

Option A:

3,216,549,870

3816547290

For option A

A: 9.86460093227

B: 9.86460093227

C: 9.86460093227

Option B = 102

Option B:

6

B-314

answer A: 3816547290

answer B: 314

answer C: 3.1415926535897932

Option B = 4

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.

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.

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

Option A : 2013456789

for A= 3816547290

option A: 3816547290

Option A: 120456045

option A 3816547290

Option A – 3816547290

Option b is room 001

for A i got 0.31

A) 3816547290

Option A is 3816547290

answer to A=3816547290

Option c is 15

Option A – 9

THE ANSWER TO OPTION a IS 1836547290

Option “a” – 3816547290

Option “b”- 0

Option “c”- 6

B) Answer is 100

Answer to problem A is:

1234567890

option c is π

Option A: 3816547290

Option a is 3816547290

Option b is 314

Option c is 3.1415926535897932

The answer to Question A is 3816547290

Option A is

I think

1232561610

A- 1234759680

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

Option A:3816547290

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.

Option A: 4

Option B: 1

Option C: 3

3816547290 answer: A

Problem A) 3816547290

Problem B) 314

Problem C) 6 (multiples)

Question A: 3816547290

Question B: 314

Question C: 3.14

Problem A: the answer is 3816547290

3

problem c : 3

Option A answer is 3816547290

C. 6

Problem A:

3816547290

Problem B:

314

Problem C:

6

3816547290

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

Problem A: 9817524360

Problem A correction: 9817524360

Final Correction: for Problem A: 9417825360

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.

Question B

Answer is room 2

answer to a: 3141592653

MATH IS NOT REAL.

WE ARE NOT REAL.

WHAT IS LIFE?

Option A = 3816547290

option c: 7

Option B: 32

Option A = 3 816 547 290

For Option A, answer is 1834567290

Answer to B is 4

Option C= 3

The answer to A: 3816547290

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

Option C: any multiple of 2 starting with 6

Option A: 50967832

Option A: 5096783214

3816547290 – Problem A

Option A; 9274561830

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

Option A: 3816547290

A. 3816547290

B. 314

C. 3.141592654

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

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

Answer to a:

3816547290

Answer A is 9,187,653,240

Option A=3816547290

Option B=21

Option C= 5

Answer to B is 6

problem A: 3816547290

Option A: 3816547290

Option B is 312

Option C: 7

A is 3816547290

The answer to the problem B is 36.

Option C answer is: 5

Option A: 1254763890 n=2

bam

Option B room # is 314

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

Answer to A: 3816547290

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?

Answer A: 3816547290

Option A: 3816547290

A . 3816547290

B . 314

C . 3.141592653589

Option A . 3816547290

Option B . 314

Option C . 3.141592653589

A: 3816547290

Option C= pie

3816547290 answer to question A

A: 3816547290

3816547290 for A

Option B:

Room 0

Option A=3816547290

Option B=21

Option C= 5

A) 3816547290

B) 314

C) 3.1415926535897932

the answer to problem A is 1296547830

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.

OPTION A: 3216748950

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.

3816547290

A:3816547290

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

Option A: 0123456789

3816547290 for problem A

C: 6,12,18,24

I Say option A: is go fuck yourself

Option B: is i lit your baby on fire

Option C: this is a setup

option B= room 314

option A 3816547290

The FIRST n for Option A is : 1234568160

Oops… copied the wrong value: Option A: 1234759680

Option A…..1234567890

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

Option B – 0

option C:5

Option A: 3816547290

Option A:

3816547290

A: 3816547290

Option A=3816547290

Option B=314

Option C=3.141592653589793238462643383279502884197169399375105820974944592307

Option A is 3816547290

3816547290 for option A

A. 987, 456, 3210

1472583690 also works for option a, so the answer is we need more information.

For Option A: 1234567890

A is 2520

3816547290 for Problem A.

Answer A is 3816547290

Problem B: 314

problem 1: 1234567890

Option C: 4.

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

Option A: 3816547290

Option A: 9,738,561,240

Option A: 3816547290

1112222520 for option c

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.

Option A:

3,816,547,290

Answer A

1234567890

option c:

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

Option C

Answer is 3

Question A: 3816547290

Answer to C: 17

2 sets of pairs =8

3 sets of triples=9

(math never my strong point lol)

A. 3816547290

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.

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

Option B: 22

Oops I read Option A wrong.

Question A: 3816547290

Question B: 314

Question C: 3.14

3816547290 Option A

option a

9147528360

option a= 1234567890

Problem A answer: 3816547290

DANK MEMES

OPTION A

123456789

N=1

Option A: 1584923760

3816547290 for Problem A

My answer to Question B is Room 314

A: a very big number

B: somewhere in between 1 and 10,000

C: 4, obviously

Option A: 3186547290

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

The answer for Option A is 3816547290

Answer to option B is room 314

Problem A:

3816547290

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

its 69 f44g0t

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

Answer to c: 9

A: 3

B:1

C:4

Answer B: room 314

Option A: 3816547290

Option A:3816547290

Option A: 9,738,561,240

Option A: 9,876,543,040

Option A: 3816547290

Option A: 3816547290

Option c: is pie which is 3.14

Option A: 3816547290

Option B: 8

Option A: 3816547290

Option A

Answer: 1234567890

Option A

3816547290

Option C: 10

The answer for A is1.9364646.93745

B is .413

c is. 763534687.e376300

Option a is 3816547290

3816547290

Option A: 3816547290

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

The Answer to Question C

Is 18

Option A is 1111111111

For option A: 3816547290 is the answer

Question A: 3816547290

Question B: 314

Question C: 3.14

option C: 11

Option a answer 20

Option C is; 3.14159… Pi

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

A 3816547290

B room 0

C 5

Option A: 1236543210

Thanks!

option A = 3941578620

Option A solution:

7,896,541,230

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.

Option A: 3816547290

Option A answer is 3816547290

Option A: 9876543210

Option B. Answer is 3

Option A:

1234567890

Option C:

24

Option A:

1234567890

Option c 12

A: 3816547290

B: 12

C: 4

The answers to all the questions are

A: 3816547290

B: 314

C: 3.1415926535897932…

Option A: 3816547290

Option A: 3816547290

[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

38165472910 was my answer

Answer A: 3816547290

C – 3.141592653589793238462643383279502884197169399375105820974944592307

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.

Answer to C is 4

I like number 4 for option A

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

Option A: 5678913240

Option a is 3816547290

Option b is 314

Option c is 3.1415926535897932

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.).

Option B the answer is 6.

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

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?

Option A: 4398765120

Option A = 3816547290

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

The answer is 3816547290 for option a

Option A: 3816547290

Problem A: 3816547290

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).

Question C: any multiple of 4

Option A: 3816547290

Option B: 314

Option C: 3.14

May I please win. Tomorrow is my birthday!

A. 3,816,547,290

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

Problem 1: 3816547290

Option A answer is 3816547290

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

3816547290 for problem A

Option B 314

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

The answer to option A = 1472589630

Option A: 3816547290

Answer for Option C:

Any integer divisible by 3. Grouped in sets of 3

Option A: 3816547290 is the answer.

Option a: 3816547290

option a) 9843156720

Option A: 1234567890

Problem A: 3816547290

Question A: 3816547290

Question B: 314

Question C: 3.14

Problem A: 3816547290

Option A

9876543210

Option C: Any multiple of 6.

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

A: 3816547290

B: 314

C: 3.1415926535897923

Answer to Option A: 1237894560

The answer to Option A question is: 9876543210.

1472583690

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

Answer: 3816547290

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?

new Thread(new Runnable() {

@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;

}

}

Problem A: 3816547290

Problem C: 36

1472583690 – Option A

Option A:

1896543270

9816547230

7896543210

1836547290

9876543210

3816547290

Option C: 8

7613459280 for Problem A

3,816,547,290

YO MAMA!!!

Option A: 3816547290

Option A: 3816547290

Problem A: 1,796,543,280

Option A: 3816547290

option A – 3816547290

Option A: 0123456789 and ‘n’=3

option c- 9

Option b: 0

Option A: 1234567890

Option C: 18

option A – 3816547290

Option A:

3816547290

Question b : 240

option A: 3816547290

Option A: 3816547290

Question A: 9184567230

A)3816547290

B)314

C)3.14159265… (Pi)

For Option 2:

1+2+3=1X2X3=6

Therefore room 6

Question A answer: 1296547830

OptionB : room 314

Option C: 3.14

The answer to B is 314.

The answer to C is 31.4

option A : 3816547290

Option A=3816547290

Option: 3816547290

Option: 314

Option: 3.1415926535897932……….

option a: 7891234560

Option A: 3816547290

Option A: 4998

Option B

answer- 0

3816547290 for Problem A

Option A: 3816547290

A: 3816547290

B: 314

C: Pi

A: 3816547290

Option A: 3816547290

The answer for option A is 0145926387

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.

Question 1

3816547290

Answer A). 93628410517

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

Option A: 3816547290 – Thanks!

A) 1022456789

Each digit divided by one ia itself.

B) Either not enough info or 314

C) 6, 12, 18, 24, 30

Q: C

Answer: 6

Option a is 9345678120

Problem A : 3816547290

Problem B : 314

Problem C : 3.14

Option A: Answer is 3816547290

Option a: 1234567891

Option b: 100

Option c: 5 or7

Option A: 1472583690

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

Answer A: 1234567890

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.

Answer C: 5.

For option A:

1,384,527,690

Problem A is 3816547290

Problem A: 3816547290

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.

Option A – 3,816,547,290

OPTION A: 1234567890

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

Option B = Room number 101

Problem A – 3816547290

Option A: 3816547290

Option A answer is 9123567480

Option B: 1

Option C: 6

Option B: 210

B ; you meet in the library

Option A: 3,816,547,290

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

Option A: 1237894560

option C: Answer is 5

Option A: 9468315720

A: 3816547290

OptionA —3816547290

First question: 4918235760

Second 6

Third: 5

I believe Option A is: 3816547290

option A

0123456789

Problem A : 3816547290

Problem B : 314

Problem C : 3.141592

Option A: 3962840571

A. 3816547290

B 314

C 3.1415926535897932384626433832795028841971693993751058209949445923

9837645120 for option A

option a is 3816547290

Question b: One

Option A: 3816547290

Question a

9345678120

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

Option B the answer is 1

The Answer to Option A = 0123456789

Option B 314

High School Student

Q.A= 3816547290

Option a- 3816547290

Option b- 12

Option c – 3.14 (pi)

3816547290 is B 😉

A) 3816547290

B) 314

C) 3.14

(One of these answers is not like the other)

Answer A: 3816547290

Answer B: 157

Answer C: Still unknown

Option A: 3.141592653

Option C: 3 to infinite

Option A: 3816547209

Question A: 3816547290

Question B: 314

Question C: 3.14

Option A is

3816547290

Option a is 3816547290

Option A: 3816547290

Option A: 3816547290

Option B: 314

Option A = 0123456789

Option B is room 3.14

B) Room #502

Option C is 12

answer to B

14

Option a is

1234567890

Option A

3816547290

Option A: 3816547290

Option B: 315

Option C: 17

Option A is 3816547290

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.

The answer to option B is: Room 314

optionA: 3816547290

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.

Option B = 6

1+2+3 = 6

1 *2*3=6

The answer to C is 6

A) is 1

B) is 1

C) is 3.141592653587993

Option C: The possible numbers are infinite (Undefined).

Example, never ending chain of key rings.

Option B the answer is 314

High School Student

Option A ; 3816547290

Option B : 314

Option C : 5

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.

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.

Answer for Option A is 3816547290

Option A is 3816547290

Option B is 314

Option C is 3.141592653589793238462643383279502884197169399375105820974944592307.

Now let’s see that pizza 🙂

option b: “not enough information”

1234759680

Option A;

3816547290

question b: 314

C – 11

Option A: 3816547290

Option C: 12

Option A

Answer 381657290

Option B : 1

A: 3816457290

C: 4

Answer to A: 3816547290

Answer to B: 0

Answer to C: 33

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;

matches.add(match);

}

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;

newMatches.add(newMatch);

}

}

}

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;

}

}

4 problem C

Option C:

6n, where n is any positive integer

A. 3816547290

B. 12

C. 3.14

Problem C:

Infinite

Question A answer is 3816547290

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.

Option A: 3816547290

C is 3*n, where n is a positive integer.

One possibility for A is 3627548190.

A has a few solutions

The answer to C is 13

Option c. 31

Problem C: 3^n

where n is the number of triple of keyrings.

Problem A answer is 3816547290

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

Problem C: 3^n

where n is the number of triples in the keyring.

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

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

— Doron Holzer

Option C: 9 or more

3816547290

Option C: Infinite possible options.

the answer to option A is 3816547290.

Answer to option b is 314

C)136

Because 1 set of rings 1 triple and 1 double

Option C

(3.14(2)) x 3 = 18.84

18 total rings

Answer to question C: infinitive

Option C: 6 or 10

Option A = 3816547290

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

My answers:

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

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

A. 3816547290

B. 314

C. 19

Option C: Any multiple of 6.

A: 3816547290

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.

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

Is the answer 13?

Option C: 12

Question A: 1834567290

The answer to Option A: 9876543210

Option C: 18

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

Problem C is 9

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.

Option A

7836549210

C:9

Option C: 54 3^3 *2

Option c: 17

Option A:

3816547290

6 to question three

OPtion C: At least 7

Option C: infinite

Option B: 0

Option C: 81

Problem C: Any number divisible by 5

48 possible for problem C

Option C: Any even number 6 or greater!

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.)

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

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.

Option c: 12

Option B: 314

A.PI

B.PI

C.PI

Option A: 3816547290

Option B is 6 right?

Option C: 4

3

answer C-18

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).

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.

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

Option C :13

Option b: 24

option c: 24 or 42

Option C:6

C: any prime number greater than 2

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

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

Option B: Room 8

Option c: 20

A:1

C: all prime numbers greater than 2

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

Answer for C is 18

A: 3816547290

Answer for C : 9

problem A answer is 4998000000

a:0123456789

b: 0

c: 3

Option C : 33

Option C: five combinations

12

121

112

212

221

C: either 7 or 13 rings

1,023,456,789

Option C : 5,6 and 7 rings

Option C : 36 ways

Option A: 1,023,456,789

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.

Option C: every multiple of 6 greater than 12

Option C : 36 ways.

Option A: 3816547290

Question C: You have 15 rings.

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

3.1415926536

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.

Question A: 9876351240

c is 19

Option A.)

3816547290

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.

6 for c

option C answer is: 7

A. is 3816547290

B. Is 314

C. Is 4 or more rings

The answer to A:3816547290

OPTION C 3 IS THE ANSWER

option c 6,12,18,24

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

C:15, +5, etc.

question c: 9 rings

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.

Option A-987654312

Option B-314

Option C-6

Option c: 19 or 13

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

Option A: 9123567480

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

Option c is 3.1415926535897932…

Option C: The answer is 15

3816547290 #PiDay My birthday!!!

Problem C: 18 key-rings

Problem C: 8 rings

A 123

B 1234567890

C 10

3816547290 for problem A

Problem A 3816547290

Option A

3816547290

Option A is 3816547290

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

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

B: 117

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

Option C.

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

Option c is 6.

C= 6

Answer to question A is 3816547290

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

Answering Option A: 3816547290.

Option C: Any integer that is 4 or higher.

Problem A: 3816547290

Option c 6

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.

Option C:

Any multiple of 6, 12 or higher.

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

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

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.

The Answer to C

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

Question C. Answer 13

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.

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.

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.

option c: 4

C Any odd number 5 or greater

C=9

twitter@kellyohioplumb

The answer to C is

n is greater than or equal to 3

B is 314

Option C

8

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

// 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();

printf("Dead End\n");

}

}

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

}

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.

Option C: 10

The answer to question A is .) 1472583690

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.

Problem C: 18 Key—rings

Option A: 1836547290

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.

Option c: 4

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.

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

My guess for question 3 is 14.

Any odd number greater than 5

Answer to C

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

C os 6n-3

But i dont know

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

Option C: 6

Problem A: 3816547290

A:0123456789

B:48

C:9

The answer to question number3 is 56.52

Option c: Any multiple of 6

6, 12, 18, 24, ect.

Can be paired and tripled

C. Any number greater than or equal to 9

C:4

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.

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

C is 6n-3

Option C: 845

Option C : the answer is 12

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

Problem A 3816547290

option C 3.14

Option C: 32

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

Answer to C:There are 3 key rings

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.

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

question c is 6

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

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.

Option c

5 and every odd number greater than 5 would work

Problem A: 4

Problem C: 9

Option C: 3,4,5

c-9

A 3816547290

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).

Option C: 3, 4, 5

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.

Option C: There is no number satisfying these properties.

Option C: 6, 12, 18, 24…..

B=0

A- 3816547290

B- 4

C- any whole number greater than or equal to 4

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.

Question C:

Any even number greater than 4.

Question C:

Any number greater than 4.

c:5

C: 4

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

Answer for Problem/Option C: Infinite Amount.

A 3816547290

B 314

C 6

Option c is 9

3 distinct 3’s but no distinct 2’s because it’s not an even number.

Option A. The answer is 3816547290.

Options C is 69 rings. No joke.

question #1 is 3

question #2 is 1

question#3 is 4

question 3/C

answer is 6

option C: 12,15,18,21 etc

Answer A: 3816547290

Answer B: 0

Answer C: Any whole number between 0 and infinity.

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

Option A: 3816547290

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

Option C: 6

answer to C. 7,8,11,12,15,16,19,20,23,24,27,28,31,32,35,36,39,40,43….

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!

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

Option C: 3 + 6x

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.

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.

Option A:3816547290

Option B:0

Option C:6

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.

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.

Problem c : 7

Option A: 1234567890

Option C is 18

Any odd number 5 or greater.

Option C: Only 5 () () ()

() ()

Option C 8

Question C: Any odd number 5 or greater.

A. 3816547290

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.

A: 3816547290

B: 314

C: 3.14

4 arranged as a pyramid

answer to C. 7,10,13,16

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.

The answer to a is 3816547290

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

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

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

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

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

Option B: 12

option c:15 rings

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

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!

Option b 101

Problem A:38165472

Answer for C:

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 ?!?)

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

Option C: Infinite number

option C: multiples of 12

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.)

Option a:

9876543210

C answer is 34

Option C: 7 rings

C ‘sanswer is 34

Answer to A: 3816547290

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

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

Answer for c is 52

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

Answer C is 70

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

Any prime number for question C

Option C: Any multiple of 10

Answer to question C:

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 .

Option C : 2

Option A is 3816547290.

12,14,28

Option C: Infinity!

See link to solution 1 here: https://onedrive.live.com/redir?resid=FA1C021FF3551BF3!8479&authkey=!AHkVFi0yvWh4D_g&ithint=file%2cxlsx

The answer is 13 for c

The answer to option C is 11 rings

Answer C is 36

For problem B, you’re in room 4

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.

Answer to Problem C:

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

…………

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

Problem A:

3816547290

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

answer to c: π

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.

Option A: 3816547290.

c=minimum of 4 to infinity

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

Option A: 3816547290

Option B: 314

Option C: 36

3.1416

Answer Option A: 3816547290.

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

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.

Question C is 6.

C=3.1416

C is 8 because there are 8 slices in a pizza

Question C: 12,15,18,21,24…

Start with 12 then increase by increments of 3.

Option A: 1,234,759,680

Answer to A: 3816547290

12 key rings for problem c

OPTION C: 2 key rings

Option A: 9,876,351,240

Option C: 36

Problem A

3816547290

Option c: anywhere from 0 through infinity.

Option C: 18

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” .

Option A :3816547290

The answer to C is 35.

Option C: Infinity

Option C’s answer: all multiples of 6

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

Option C: Infinity

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.

option c: 12

Option c:

3x + (x-1)

Where x is the amount of triples and x cannot be less than 2

A: 3816547290

Option C: 3, 12…

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!

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.

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.

The answer to C is 35

Option C:

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

OPTION C: 2 KEY-RINGS

Option A: 3816547290

Option C.

Answer is 6… a pair of triplets.

Option C: 3.00

OPTION C IS 35

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.

Answer to C:

3 or 5

Option C

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

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.

Option C: 12

option b, 1

Option C) 3.1416

Option C – 11

Problem A: 3816547290

Option C: 7 Keyrings

Option C:10

C: 13 keys

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

C: 13 rings

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

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

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.

C: 18

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.

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.

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

The answer to Option C is 6

Answer to C: any number that is divisible by 6

C: any number that is divisible by 6

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.

C: 6, 12, 18, etc.

Option C is greater than 9

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

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.

problem A=3816547290

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.

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

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!

Greater than or equal to 5

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.

The answer to Question A is 3816547290

For question C = 413

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

A=3816547290

Option c greater than or equal to 5

Option C:

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

7836541290 for A

Problem B: 1

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

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.

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.

For Question C: 3

3816547290 for problem A

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

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

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.

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

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

answer for option A

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

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.

Attempt at question c:

Possible answer: 27

As I don’t see my original comment which should have been posted before the deadline, option c: 3 to infinity.

The number of points in a 3.14 sq. in. circle where two circles are tangent approaches infinity as we lack data to limit that. We know a triplet is observable so a minimum of 3 is determinable but the area taken up by that point of 2 tangent circles is not given. As such, it can be assumed to be infinite. There is also the matter if we are limiting ourselves to a 2D plane or a 3D sphere but that doesn’t change my answer. Like pi, this answer does not terminate and might be a little “irrational.”