## PRIZE PUZZLE FOR JUNE 2008## DIVIDED CAKETimothy Hardcastle does not like cake but he does like parties. Every birthday he invites guests to share his birthday cake. He insists that the whole of the round cake is eaten, all 360 degrees of it! His guests are allocated a prime number which determines how much of the cake each guest will have, the number being the number of degrees equal to the prime number allocated. No guest is allowed to have the same prime number as any other guest. Timothy, a mathematician, always ensures that the whole cake is eaten by making sure that the sum of the prime numbers allocated comes to 360. For example, when he had two guests he allocated the numbers 179 and 181 which was as fair as you can get (he could have had 83 and 277 but that would not have been as equitable). For three guests he could have allocated the numbers 2, 89 and 269. (Cutting 2 degrees of cake might be a bit difficult but never mind the practical details!) For ten guests he could have given the numbers 11, 13, 17, 19, 29, 37, 47, 53, 61 and 73 to his guests. The question is, what was the maximum number of guests he could invite to his party according to these rules? If you would like some help with those prime numbers, try the link Table of the first 100 Prime Numbers.
I received only two answers to this problem: they were from John Stafford and Richard Birkill. Many thanks. Richard Birkill also supplied the results of a computer search that found dozens of solutions. Here is one: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 79 There was no draw this month as there were less than three competitors. I think the mention of prime numbers put people off. |