National Pi Day Math Problems Solved!

March 15

A big thanks to all of our pizza loving math aficionados! We sent out the challenge and you answered in a big way. There were many correct answers but some typed faster than others. Winners have been notified via email, but we couldn’t bear to leave everyone hanging so answers are below!

OPTION A: 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: 3816547290 


OPTION B: 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?”

ANSWER: Room #12

The numbers in the envelope are either: 6222 or 4431, which both add up to 12 and the product is 48

  • To truly get this right, you must first eliminate that #13 is NOT an option
  • To do this you adjoin a 1 to the above numbers: 62221 or 44311, which both add to 13 and the product is 48
  • And if you take 922 and 661, which both add to 13 the product is 36

OPTION C: 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?

ANSWER: Despite the internet’s best effort this one remains UNSOLVED! See below for a note from John H. Conway himself:

John Conway - Pizza Hut - Answer to Hardest Math Problem

It reads: “If Any Solvers find this answer (which I’ve kept a closely guarded secret). I’ll look at their answers myself (there won’t be many perhaps even none).

Right again Professor Conway.

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  • Joe Bohanon March 15 at 4:47 pm

    It’d be nice if we could get a translation of the Option C problem into math language from Professor Conway. I was almost sure that my answer of classifying doubly-transitive, but not triply-transitive permutation groups, was correct.

    • David March 16 at 12:19 am

      That’s a necessary condition (although I think it’s a bit ambiguous whether we care about ordered versus unordered “pairs”/”triples”). But there’s still topology remaining! Namely, one has to determine which of those two-transitive, non-three-transitive groups actually arise as the “symmetry group” (appropriately defined) of a jumble.

      And that’s a nontrivial task, since already for three rings there are nutty jumbles like the Borromean rings. Probably it’s important to also use our geometric knowledge: the key rings are almost the same size and almost circular. So many constructions that’d be possible with loops of string are disallowed (e.g. two rings are either linked or unlinked, not wound around each other twice, thrice, etc.).

  • Liam March 15 at 4:50 pm

    Answer to C: 9

  • Rachel Toldoya March 15 at 4:57 pm

    There are at least 2 sets of key rings. On each key ring there are at least 3 rings. As far as I can tell that means you must have at a minimum 6 key rings. It could go anywhere up to and including

    ______|______|______|______|______|______|______o_>
    0 1 2 3 4 5 6

  • Justin March 15 at 5:11 pm

    The answer to the last question is 1.

  • Stephanie March 15 at 5:14 pm

    Could someone please explain how you knew what the numbers in the envelope were? Why did it have to multiply to 48 and add to 12?

    Also, could some please explain how you got the number for option a.

    • Lisa March 15 at 9:44 pm

      The girl knows the number of the room in which they meet. She asked whether she would be able to determine what numbers were in the envelope if she was told the number of numbers in the envelope, and the product of those numbers. The answer to this question was “no.” Therefore, the room number must be one that can be split up into parts (such as 12 being split up into 6, 2, 2, and 2) such that there are two ways you can split it up resulting in two different sets of numbers that meet the following requirements:

      1. Both sets add up to the room number.
      2. Both sets contain the same number of numbers.
      3. Both sets have the same product when multiplied.

      The first possible number where this is possible is 12. The two sets of numbers that meet the above requirements are (6, 2, 2, 2) and (4, 4, 3, 1). Both of those sets add up to 12, have 4 numbers, and have 48 as their product. The girl was able to determine the product of the numbers in the envelope, without knowing the numbers. The only way that would be possible is if she knew the number of the room (which she did) and knew that there were only two possible sets of numbers that met the above requirements. Since they both have the same product, she was able to tell what the product was without knowing which of the sets of numbers it was.

      Since she was told that she would not be able to determine the set of numbers, she knew that the set MUST be one of the sets that shares its product with another set, otherwise she would be able to tell which set it was based on the number of numbers in that set and the product of those numbers. The answer cannot be anything above 12, because starting with 13, there are too many ways to divide up the number and still meet those requirements. If it were a number that was higher than 12, she would not have been able to figure out the product, because she would know that there was more than one possible answer. I hope that makes sense.

  • Paul Pence March 15 at 5:23 pm

    Here’s my answer for C

    This answer is in retaliation for the question not defining what “pairs” and “triples” are. It is based on the idea that there are two kinds triples — those made by three rings laying side by side (T), going through the same rings as each other, and those rings which have exactly 3 other rings going through them (t), no matter how those rings subsequently attach to other rings. Similarly, there are pairs of side-by-side rings (P) and rings with two other rings passing through them (p).

    With these definitions, we can tell some triples (T) from other triples (t), if both kinds of triples exist in the construct.

    We can also not tell the difference between pairs if all of the pairs are of the same variety.

    The last factor to consider is the inability to distinguish left from right. I take this to be a universal statement, so front/back, top/bottom, and clockwise/counterclockwise can not be determined. If that is the nature of the construct, it requires symmetry in the construct.

    The smallest construct with both types of triples is where three rings lie on top of each other with a single ring going through the stack of 3. Stacked triples (T) = 1 and Loose Triples (t) = 1. The number of pairs p = 0. N=4

    A single ring can be added, going through the 3 stacked rings, but not through the unstacked ring. This makes T=1, t=2, and p=0. N=5

    In the same manner, a single ring at a time can be added, creating additional unstacked triple (t), with N increasing by one for each one added.

    In the theoretical sense, solutions occur for all N from 4 to infinity, but there are practical limits.

    At some point, it would be necessary to add a ring to one of the unstacked rings. For symmetry to be maintained, a minimum of two rings must be added at opposite ends of the construct, creating unstacked pairs. At any time, do not add stacked pairs, since then you would be creating a kind of pair that could be differentiated from the other kinds.

    So the answer, using definitions I felt were left up to me, is N = 4 to infinity.

    • Brian March 16 at 8:45 am

      Didn’t finish reading your solution, but note that 5 can’t possibly work: If two of the triples of rings are distinguishable (say ABC and ADE), then the pairs of rings that aren’t used (DE and BC) are also distinguishable.

  • alex March 15 at 6:05 pm

    OPTION C: 5040

    this is so weird i went to bell ringing

  • Vannesa Nalle March 15 at 6:06 pm

    the answer is 27inches in diamiter

  • Brandon Guilding March 15 at 6:07 pm

    I think for c that the answer is 4,5,6…. any natural number larger than 3.
    When we draw 3 points and connect them with an edge, then we add a fourth point, this gives us 2 different sub graphs that have 3 key rings.
    If we add another point we continue to have different sub graphs with 3 and 2, so we need at least 4 nodes in our graph to meet the criteria.
    I almost though about using Ramsey theory to answer the question. For example if some triples are distinguishable then we want a number of nodes such that we get at least 2 coloring’s of a sub-graph that contain 3 edges of a color that are different from the other edges in the graph.
    and the rest of the edges can be a color such that they are indistinguishable. We color our triples that are distinguishable green, and any other edge say red, we cant know right from left because our graph can be rotated and be equivalent.
    So the answer is the ramsey number for R(n3,2) where n is a multiple of 3, this will give us more than 1 sub graph.

  • Bubba March 15 at 6:26 pm

    Option C. 40

  • bubba March 15 at 6:32 pm

    option c 40

  • bubba March 15 at 6:33 pm

    option c: 13

  • Timothy Brown March 15 at 7:12 pm

    tried posting my answers for option c to no avail. option c is N^2 + N(N-1) where N > 1. 6, 15, 28, 45, 66, 91 etc. This would produce hexagons with sides of N links when assembled as 1,2,3,4,4,4,4,3,2,1 where N = 4 in this example. Alternatively I also considered, the option to add 1 ring in center of any 4 linked rings in the previous equation, which would produce an equation of N^2 + N(N-1) + (N-1)^2 where N > 1. The first answer, is 7, 19, 37, 61, 91 etc.

  • Bubba March 15 at 7:13 pm

    option c: 13

  • Bubba March 15 at 7:14 pm

    Option c: 40

  • Zoila Martinez March 15 at 7:14 pm

    I´ve got an answer.
    Problem C: 18 key-rings

  • magnum godfrey March 15 at 7:26 pm

    c= minimum 3 rings and any amount thereafter, because of the jumble. they are all tied to one ring.

  • Jaime Vazquez March 15 at 7:33 pm

    There are six key rings.

  • Ryan pacheco March 15 at 7:48 pm

    How can option 1 be correct? 3 does not divide 290.

    • Alec March 15 at 9:56 pm

      Start from the left: 3 divides 381

    • k March 16 at 12:30 am

      I believe, “the first n of them,” means the integers from left to right, not right to left. Therefore:
      n= 1: 3/1=3
      n=2: 38/2=19
      n=3: 381/3=127
      n=4: 3816/4=954
      n=5: 38165/5=7633
      n=6: 381654/6=63609
      n=7: 3816547/7=545221
      n=8: 38165472/8=4770684
      n=9: 381654729/9=42406081
      n=10: 3816547290/10=381654729
      I hope that clarifies things for you Ryan.

    • Sarah March 16 at 7:51 am

      By “first n digits” he means left to right. 3 is divisible by 1, 38 is divisible by 2, 381 is divisible by 3, etc.

    • Nick March 16 at 8:44 am

      They mean “first” as in left to right — so the first three numbers are 381.

    • Liam Landrum March 16 at 9:07 am

      you are thinking of it in reverse, the first digit it is 3 and then 38, then 381 etc.

  • Brian Stephens March 15 at 7:52 pm

    For problem B, I agree that Room 12 is a possible answer, but is it the only one? My answer was Room 16:
    The numbers could be 2,5,9 or 3,3,10, which are both sets of 3 numbers adding up to 16 and having products of 90. So if the number-hider told them just the number of numbers and their product, that would not be enough to determine the set of numbers.

    • Mitch March 16 at 5:14 am

      16 is wrong for the same reason that 13 is wrong (and every other number larger than 12, for that matter. Because one of the puzzle club members know what the product was, there could in fact only be one unique product for the solution. For every number greater than 12, there is a solution that includes the two sets of numbers that made up the solution for 12, and enough additional ones to get to the number in question which both have products of 24. For example, for 16 it would be 4,3,2,2,1,1,1,1,1 and 6,2,2,1,1,1,1,1,1, Therefore 12 is the only number that has a unique product solution.

    • Robin March 16 at 7:29 am

      Problem B was poorly worded. Instead of a “list of numbers” in the envelope, it should have said something like “sequence of digits”.

      A “list of numbers” could mean “-3, 4.25, 1.76”. There was no indication that it was restricted to single-digit numbers, nor positive, nor integer. And though the problem was unsolvable without those three restrictions, perhaps it’s not impossible to come up with a different set of assumed restrictions under which the problem would work.

  • Emily DiDonato March 15 at 7:59 pm

    Here’s an odd idea for C. Suppose there are 2 separate kinds of rings. For instance, 2 different colors, A and B. These rings are locked together in one large, straight loop of alternating colors. Then, every pair would indistinguishable (A-B, with A connected to another B, and B connected to another A). But, there would be 2 types of triples: A-B-A and B-A-B. In this case, the answer would be all even numbers greater than or equal to 4. (I realize it’s a little late, though, and that it makes some odd assumptions about the problem.)

    • Bob March 15 at 10:21 pm

      I completely agree with your analysis and wrote that but I haven’t received the prize yet!

  • Reece McDevitt March 15 at 8:03 pm

    I think the answer to problem #3 is any triangular number that is greater than or equal to 3. If you hold the ring of key chains taut, you can make them into a triangular network (with each triangle being equilateral with a side length of 2 since each radius is 1). You can’t tell pairs from each other (since the chains are held together in triangular clusters), however, you can tell the triples. From the vertices, you can tell the triples from the outside (the two chains that are adjacent to each vertex) and then you can just work your way inside using adjacent rings to tell your triples. A diagram would help a lot, but unfortunately I can’t upload a pic!

    I hope I’m at least somewhat close. This was such a cool problem for me to tackle after work today! Very invigorating. Hope to see the answer 🙂

    -Your amateur Math enthusiast

    • Anthony Romano March 16 at 8:02 am

      Option C: My answer is based on the theory that all fullerenes have exactly 12 pentagons. The smallest fullerene has 12 pentagons and zero hexagons. You can’t make a fullerene with 1 hexagon, so that will be skipped. All other values greater than or equal to 2 hexagons are possible. I’m calculating the number of rings to be the number of edges on the fullerene. Each vertex having 3 connecting edges.

      Formula: v = vertices, e = edges, f = faces, p = # of pentagons, h = # of hexagons
      e = v – 2 + f
      e = (5p + 6h)/3 – 2 + (p +h)

      Where p=12 and h is all values >= 0, excluding 1. The possible number of rings is as follows: starting with 30, (33 not possible), 36, 39, 42, 45, etc. The sequence continues adding 3 more rings each time.

  • Josh March 15 at 8:09 pm

    Option c answer: 2

  • pharenna pol March 15 at 8:15 pm

    answer to option c – multiples of 3 but not divisible by 2.

    • Travis March 16 at 2:27 am

      A really big issue with option C is the fact that pairs and triplets can be taken very very differently based on different people. A viable answer for C is the number 4. Just 4 interlocking ones. This sets up a square in which no pair is totally distinguishable and also the triplets are because they would all form triangles regardless of which set you chose. Thus any form of that square would work in a way that would technically solve this problem, no matter how big it becomes. That also would settle being unable to tell which way comes first because both sides would be even.

  • Ed C March 15 at 8:33 pm

    Option C: The possible numbers are 5, 9, 13, 17, 21, 25, then 25 +4 to infinity if there is no limit. The figure is a central circle (key-ring) with four around (totaling 5), then adding 4 at the intersection of each 4, totaling 9, etc… Each of the four initial circles are joined together 1 inch (half way). The radius will force them to be at 90 degrees (say forming a position in the 12, 3, 6 and 9 O’Clock positions) then the rest of them radiating out forming another 4 circles at the 1:30, 4:30, 7:30 and 10:30 positions. The possible design will be limited to the 2 inch diameter times PI which will define the radiating figure.

  • Jalisa March 15 at 8:35 pm

    Question C) 15

  • Chris Boehle March 15 at 8:35 pm

    C: multiples of 9

  • alex March 15 at 8:58 pm

    10, 792

  • Ed C March 15 at 9:28 pm

    Option C: The possible numbers are 5, 9, 13, 17, 21, 25, then 25 +4 to infinity if there is no limit. The figure is a central circle (key-ring) with four around (totaling 5), then adding 4 at the intersection of each 4, totaling 9, etc… Each of the four initial circles are joined together 1 inch (half way). The radius will force them to be at 90 degrees (say forming a position in the 12, 3, 6 and 9 O’clock positions) then the rest of them radiating out forming another 4 circles at the 1:30, 4:30, 7:30 and 10:30 positions. The possible design will be limited to the 2 inch diameter times PI which will define the radiating figure to extend with 4 additional rings outwards.

  • Tre March 15 at 9:32 pm

    C=9

  • Dontreveon March 15 at 9:37 pm

    The answer to C is 4.

  • AW March 15 at 9:46 pm

    Dr. Conway, I apologize if I am venturing in a completely and embarrassingly wrong direction in your field, but Option C seems to be to be about knot theory (“strange jumble”), and given the statement about right vs left, we would seem to have a hemispheric-ly symmetrical knot, where links of three (Borromean rings?) are discernible. The diameter is relevant to considering the circumference and deriving from there the maximum size of the knot, which, assuming it’s three-dimensional, means considering the volume of a sphere, unless the left-right mixup really indicates a knot relatively flat on the plane. Those are just my hunches, and regardless of whether I am wrong in the most insipid way, the contemplation has been fun. Thanks, Dr. Conway and Pizza Hut, for Pi Day joy!

  • Andrew Sorensen March 15 at 9:48 pm

    How do we know the set of numbers or the product for B? the question does not state either.

  • Alec March 15 at 9:59 pm

    I still think the answer to C is 4 or more. I think the problem needs to be state more clearly.

    One thought was that connected pairs could be distinguished from disconnected pairs of rings, meaning the first statement implies every ring must be connected to every other ring., and it would depend on some physical limit, but then I’m not sure how would you distinguish triples?

  • Matthew Shaw March 15 at 10:01 pm

    So, did we miss out or is the prize still available?

  • Ed C March 15 at 10:02 pm

    Option C: The possible numbers are 5, 9, 13, 17, 21, 25, then 25 +4 to infinity if there is no limit. The figure is a central circle (key-ring) with four around (totaling 5), then adding 4 at the intersection of each 4, totaling 9, etc… Each of the four initial circles are joined together 1 inch (half way). The radius will force them to be at 90 degrees (say forming a position in the 12, 3, 6 and 9 O’clock positions) then the rest of them radiating out forming another 4 circles at the 1:30, 4:30, 7:30 and 10:30 positions. The possible design will be limited to the 2 inch diameter times PI which will define the radiating figure to extend with 4 additional rings outwards. That is, the diameter limits the extension of the radiating, linked circles.

  • Anthony Romano March 15 at 10:03 pm

    Option C: I’m going with the theory that all fullerenes have exactly 12 pentagons. The smallest fullerene has 12 pentagons and zero hexagons. You can’t make a fullerene with 1 hexagon, so that will be skipped. All other values greater than or equal to 2 hexagons are possible. I’m calculating the number of rings to be the number of edges on the fullerene.

    Formula: v = vertices, e = edges, f = faces, p = # of pentagons, h = # of hexagons
    e = v – 2 + f
    e = (5p + 6h)/3 – 2 + (p +h)

    Where p=12 and h is all values >= 0, excluding 1. The possible number of rings is as follows: starting with 30, (33 not possible), 36, 39, 42, 45, etc. The sequence continues adding 3 more rings each time.

  • Andrew Sorensen March 15 at 10:03 pm

    Could someone explain B?
    is it supposed to be read that the product/number of numbers is also equal to the sum of numbers?

    or could i see a written explanation as to why the answer is 12?

  • Chris March 15 at 10:04 pm

    Option C: n^2+4n for all positive integers greater than 1

  • Marty Hensley March 15 at 10:09 pm

    I used to be a custodian at a school and I ran into this exact problem with my keys. I had one ring that was connected to two rings that were connected to another ring each bringing the total to five rings. Sometimes depending on the events I would need to carry another set of keys along with these five , this other set had three rings all connected to each other and I would connect them to the other five to make one large jumble of keys , and I could never tell any pair from another pair but some triple were like other triples and there was no chance of telling left from right anywhere, I just had to know the keys by the color of the head cover on them. So I’ll say the answer can be 5 & 8 , and then 11, 14 and so forth

  • Tre March 15 at 10:16 pm

    C = 3 or 4

  • Andrew Sorensen March 15 at 10:16 pm

    Option C:
    the minimum number of rings is 9.

    ring 1 is connected to 2, 4
    ring 2 connected to 1,3,5
    ring 3 connected to 3,6
    ring 4 connected to 1,7
    ring 5 connected to 2,8
    ring 6 connected to 3,9
    ring 7 connected to 4, 8
    ring 8 connected to 7,5,9
    ring 9 connected to 8,6

  • Dontreveon March 15 at 10:17 pm

    C is 9

  • Drake March 15 at 10:18 pm

    The answer is 4

  • TJ March 15 at 10:20 pm

    4 is C

  • Cdickey March 15 at 10:23 pm

    The answer to C is zero.

  • Jonathan March 15 at 10:45 pm

    C. Any whole number greater than 4

    • Jonathan March 15 at 10:51 pm

      Edit: Any whole number great than or equal to 4

  • AW March 15 at 11:01 pm

    Option C: 39
    (?)
    I had previously posted about knot theory, but I was probably hasty in blundering towards a guess, as I am still and now. Is this really about Leech lattices and sphere packing? If the strange jumble is really an omni-directional icosahedron with triples (Borommean rings) substituting for spheres, then it would be (12 spheres x 3 rings comprising each sphere) plus 1 triple at center = 39

    It’s fun to contemplate, even if in a wrongheaded way.

  • Jon S March 16 at 12:41 am

    The answer to option C is any multiple of 3 equal to or greater than 9. Being able to distinguish triplets but not pairs indicates that each ring must be connected to two others. If any ring were connected to only one other it would create the ability to distinguish left and right as well. Some might argue that it also cannot be any number divisible by two, but that isn’t quite the case as some of the triplets may have more superfluous connections to each other, which would be why you’re describing it as an asymmetrical jumble.

  • Tom C March 16 at 1:03 am

    Option C: Since my first answer wasn’t right, I re-examined my assumptions based on the language of the problem. It still appears to indicate at least two doubles and two triples. The most basic “tangle” would be the two doubles looping through a triple on either side with a single looping through each of the doubles to make them separate (1-2-3-2-1). The second triple also loops through the first triple giving you two of each but making the triples distinct. Other singles and triples could be added but any key-rings connecting to the doubles would have to be the same for each of the doubles. More doubles could be added but they would have to loop through the first triple and have the same additions as the other doubles.

  • Steve March 16 at 1:25 am

    Option C:
    Key Rings = 3(2^n – 1) for any integer n >=1

  • Ryan March 16 at 1:52 am

    For problem C, all integers >=5.

  • Rather annoyed. March 16 at 2:17 am

    So the correct answer to C is that it doesn’t exist. There is no proper answer. They even failed to post the answer.

  • Travis March 16 at 2:29 am

    A really big issue with option C is the fact that pairs and triplets can be taken very very differently based on different people. A viable answer for C is the number 4. Just 4 interlocking ones. This sets up a square in which no pair is totally distinguishable and also the triplets are because they would all form different triangles regardless of which set you chose. Thus any form of that square would work in a way that would technically solve this problem, no matter how big it becomes. That also would settle being unable to tell which way comes first because both sides would be even.

  • Heather March 16 at 4:44 am

    a=4(pie)r^2
    12.56 or rounding up 13 but I think this is only one step, I believe that there is “knot theory ” involved with the answer.

  • Mitch March 16 at 5:00 am

    I believe 1472583690 is another solution to A. And it’s arguably more elegant as it resembles sequences on a numeric keypad or telephone keypad.

    As far as problem C, I believe the correct answer is that the possible number of key rings in this jumble are 6 or 12. These are the number of sides in all of the regular convex hedrons that do not have triangular faces. The key rings would be configured as though they were the faces of these geometric shapes. I think that the notion that, “I can’t tell any pair from any other pair” implies not only the relationship that the rings in the pair have to each other, but also the relationship they have to every other ring around them. What is interesting about all convex hedrons is that any two adjacent sides have the same spacial relationship with each other and with all the remaining sides, regardless of which two adjacent sides you choose to be your initial point of reference. Because all the key rings are the same size, we must restrict the universe of hedrons to regular ones (i.e. ones where all the faces are of identical size and shape, and edges are of identical length, which takes us down to 5 shapes (or key ring configurations. Note also that all non-convex regular hedrons have faces that are isosceles triangles and therefore would not be suitable surfaces for the circular key rings to reside upon). Of the five shapes that remain, three of them: the tetrahedron (4 faces), octahedron (8 faces), and icosahedron (20 faces) all have equilateral triangles as their faces. If we consider any 3 adjacent sides of these objects, they will have the same spacial relationships to each other, and to all the other faces (keyrings), regardless of which 3 we choose. Therefore, they do not conform to the statement that, “I can tell some triple from other triples”. Only the cube (6 faces) and the dodecahedron (12 faces) have triples (three adjacent sides) that can be chosen such that they have more than one unique spacial relationship with each other. For example, pick up a common die and note that the sides that contain 1, 2, and 3 dots are all adjacent. Then look at the sides that contain 1, 2, and 6. Again these are adjacent, but they have a very unique spacial pattern, i.e. the triples are distinguishable from each other. Thus the answer to the problem is that there may be 6 or 12 key rings in the jumble.

    • Mitch March 16 at 6:23 am

      Never mind about the solution to A. I got careless and missed n=8 doesn’t work.

  • Anthony Romano March 16 at 7:52 am

    Option C: My answer is based on the theory that all fullerenes have exactly 12 pentagons. The smallest fullerene has 12 pentagons and zero hexagons. You can’t make a fullerene with 1 hexagon, so that will be skipped. All other values greater than or equal to 2 hexagons are possible. I’m calculating the number of rings to be the number of edges on the fullerene. Each vertex having 3 connecting edges.

    Formula: v = vertices, e = edges, f = faces, p = # of pentagons, h = # of hexagons
    e = v – 2 + f
    e = (5p + 6h)/3 – 2 + (p +h)

    Where p=12 and h is all values >= 0, excluding 1. The possible number of rings is as follows: starting with 30, (33 not possible), 36, 39, 42, 45, etc. The sequence continues adding 3 more rings each time.

  • Andrew Sorensen March 16 at 7:59 am

    Option C: 9 key rings

  • Tre March 16 at 8:02 am

    It could be 4,5,6 or 9

  • Andrew Sorensen March 16 at 8:02 am

    Can someone explain why b is 12?
    we are not told the value of the product and we can’t assume that the product/number of numbers=sum. so how do we determine the numbers?

  • Jeff March 16 at 8:29 am

    Option B: But Room #12 could also be the following sets of indistinguishable cards: 3, 3, 2, 2, 2 or 18, -2, -2, -1, -1. Five cards that have a product of 72. So if they are in fact in Room #12, the girl still didn’t have enough information to conclude the product. But for that matter, the numbers could have easily been 12, 0, 0 or 6, 6, 0 (three cards with a product of 0). Some commentators concluded that because he took time to scribble some things before answering, that means there wasn’t a zero card (because he could have answered that question easily otherwise). But who’s to say he wasn’t being deliberately deceptive? Nonetheless, if the numbers could be negative (when was it ever specified they couldn’t be?) and if we are meant to conclude there isn’t a zero by the scribbling involved, I’ve just shown ambiguity in the answer.

  • Bethanie March 16 at 8:29 am

    I dont understand why the answer for option B cannot be room 14. for example, the numbers 8,3,3 and 6,6,2 both have a sum of 14 and product of 72.

  • Kristi March 16 at 8:37 am

    The least amount it could be is 10. After, it would be multiples of 10

  • Nick Mangum March 16 at 8:43 am

    Could you please give explanation or re-word Option C since no one has solved it? I also feel like designing this blog so that anyone can read the submissions shoots a massive hole in the entire process…

  • Kristi March 16 at 8:45 am

    Option C: the least amount it could be is 10. After it would be multiples of 10

  • Tony Bianco March 16 at 8:49 am

    Possible solution to Problem C: Dr. Conway, when I initially considered this problem I immediately thought in terms of group theory (which is not my strong point). On the other hand, “pairs” and “triples” lead us to multiples of two and three respectively. Both numbers are prime and have LCM 6. The prime factorization for any multiple of 6 is (2^n)(3^m) where n and m are integers greater than or equal to 1. So, a multiple of 6 can be thought of as a “pair” of “triples” (or a “triple” of “pairs”). Since we can only tell some “triple” from other “triples”, we need to consider the odd multiples of 3 (the multiples of 3 that are not also multiples of 6). In other words, the possible number of rings is 3 mod 6 (or remainder 3 when divided by 6). So,
    r=3, 9, 15, 21,…

  • Ethan F March 16 at 8:53 am

    Answer to C: Any integer >= 5.

    I’m making some assumptions about the wording of the problem here:
    – A triple is a connection of exactly three rings, and a pair is a connection of exactly two rings, so a triple does not consist of three pairs.
    – There must be at least two triples, since the problem suggests that there exists triples A and B such that A and B are distinguishable.
    – There can be any number of pairs (including 0 or 1), since those still satisfy the requirement that for every two pairs A and B, A and B are not distinguishable.

    The minimum number of rings to create two triples is 4. Take one triple of 3 and link another ring to two of the three. This creates two links of exactly three rings, and zero links of exactly two. The only issue is that the two triples are not distinguishable, so one more ring must be added on one side in order to create a way to tell the triples apart. This introduces the only pair, which is fine since there are no other pairs to compare to. So, we can satisfy the requirement of the problem with 5 rings.

    Now, using the method that we used before to create the initial two triples, we can attach rings one at a time to two of the rings in one of the outermost triples. Each ring introduces a new triple but no new pairs, so we can safely repeat this process without breaking the constraints of the problem. This means that there can be any whole number of rings greater than or equal to 5.

  • Liam Landrum March 16 at 9:06 am

    Option C: based on the info given, it seems that the answer to this question could be theoretically 4 to infinity as you need at least four rings to have two different sets of three ( assuming that these triples overlap. There is also the possibility that there is no answer based on each pair of two having to be the same but each set of three to be different. Sadly the vagueness of this question cant help us understand what is a triple and what is a pair and if they can overlap or must be distinct.

    Option B: I would like someone to explain to me how 12 is this answer. Theoretically the answer could be any number as adding 1 does not change the product but would change the sum, hence being unable to find the room number. If 0 was in the envelope the room number wouldn’t change because zero wouldn’t change the sum, but it would always set the product to zero. Maybe I’m thinking in depth to much but I would really like some explanation on this.

  • Jared Rudolph March 16 at 10:17 am

    Maybe it is odd numbers greater than seven