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 Prime Perimeter (Posted on 2005-03-11)

You and two other people have numbers written on your foreheads. You are all told that the three numbers are primes and that they form the sides of a triangle with a prime perimeter. You see 5 and 7 on the other two heads and both of the other people agree that they cannot deduce the number on their own foreheads.

 See The Solution Submitted by Erik O. Rating: 2.8235 (17 votes)

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 Trying to solve this... | Comment 35 of 90 |

Wow, so many comments!  I sort of glanced over all of them, and I didn't see any satisfactory solutions.  I don't agree with pcbouhid's solution, but I didn't look long enough to be sure.  In any case, I'd rather just solve it myself.

If you're not interested in reading this whole thing, skip ahead to the final thoughts.

My interpretation:

I'd rather call the three people A, B, and C, with C referring to "you", and A the person with a 5.  It is rather frustrating and confusing to use second and first person.

"Agree" means that A and B each say that they cannot determine the number on their own foreheads.  They will not be able to determine their own numbers no matter how much each declares. it.  It is like one of those logic puzzles where A says, "I don't know my number," and B says "Me neither."  A says, "I still don't know," and B says "Me neither."  They can continue indefinitely without figuring it out.  C is not part of this conversation.

There are three limits on the numbers:  each is a prime, their sum is a prime, and they obey the triangle inequality.

My solution (solving as I type):

The possibilities for C's number are 5, 7, and 11.  These are all the primes between 3 and 11, excluding 3 because 3+5+7 isn't prime.  I will eliminate the possibilities one by one.

If C=7:
A thinks My number is 3 or 5.  He declares he doesn't know.
B thinks My number is 5,7, or 11. He declares he doesn't know.
A thinks If my number were a 3, B would have deduced his number.  Therefore my number is 5!  He declares his number.

If C=5:
This quickly gets complicated, so I will use the column method (as I call it).  The left-most column shows all the possibilities that I think are relevant.  Usually, when I do this, there are finite possibilities, so I hope this still works!  The following columns represent the alternating responses of A and B.  X represents figuring it out, O represents not knowing.  The underlined possibility is the true one.

`3,3,5   OO3,5,5   OO5,3,5   OO5,7,5   OOOOO...7,5,5   OO7,7,5   OO7,11,5  OO11,7,5  OO11,13,5 OO13,11,5 OO13,13,5 OO`

Uh oh... that means I can't eliminate 5.  I hope I can eliminate 11.

If C=11:

`5,7,11   OOOOO...5,13,11  OO7,5,11   OO7,13,11  OO11,7,11  OO11,19,11 OO13,5,11  OO13,7,11  OO13,13,11 OO13,19,11 OO`

...I guess I can't eliminate 11 either.

Final Thoughts:

I was wondering why the below comments couldn't seem to figure this one out.  Now I know why everyone's having problems--because there are two solutions: 5 and 11.  Either my solving is wrong, or my interpretation is wrong.

Edited on March 14, 2005, 12:51 am
 Posted by Tristan on 2005-03-13 01:50:45

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