Day 8: The Gift Exchange

Algebra Level 5

A group of 1000 1000 mathematicians are participating in a Christmas gift exchange. The organizer of the gift exchange wants to add a mathematical twist to the gift exchange, so he decides that in this gift exchange, all of the mathematicians will be numbered with a number from 0 0 to 999 999 . Some positive integer 0 < n < 1000 0 \ < \ n \ < \ 1000 is chosen, and some subset of the mathematicians of size a a is also chosen. For each mathematician in the subset of the gift exchange, they must give a gift to the person with the number that is equivalent to the remainder when their own number plus n n is divided by 1000 1000 . The mathematicians giving and receiving all the gifts must be contained within the subset. How many different a a values exist in the form p t p^t , where p p is a positive prime number, and t t is a positive integer, such that the person with the number assignment 0 0 is always in the subset?


This problem is part of The 12 Days of Math-Mas 2018


The answer is 6.

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1 solution

Jack Ceroni
Dec 17, 2018

When examined further, this problem can actually be solved fairly easily using group theory. We are given a group operation, which is addition modulo 1000 1000 , and a set of integers on which to act. Since this problem tells us we must include the mathematician with the 0 0 designation within our subset of mathematicians, we can define 0 0 to be the identity element of the subgroups that we are trying to find, since only subgroups will obey the given criteria within the question (the mathematicians must give a gift to the person with the number that is equivalent to the remainder when their own number plus n n is divided by 1000 1000 , and all mathematicians giving and receiving gifts must be completely contained within the subset). By Lagrange's Theorem, we know that G H \frac{|G|}{|H|} , where G G is a group, and H H is some subgroup. The different a a values that we are trying to find are essentially the different orders of the subgroups, so we know that all a a values must divide the order of G G , which is 1000 1000 . Taking the prime factorization of 1000 1000 , we get 2 3 5 3 2^3 5^3 . All a a values in the form p t p^t that divide 2 3 5 3 2^3 5^3 are 2 1 , 2 2 , 2 3 , 5 1 , 5 2 , 5 3 {2^1, \ 2^2, \ 2^3, \ 5^1, \ 5^2, \ 5^3} , therefore the number of different a a values is 6 6 .

For p = 2 p = 2 , taking n = 500 n = 500 and ({1,501}) as the set of mathematicians seems to work, so 0 does not have to be in the set.

Jon Haussmann - 2 years, 5 months ago

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Yes, but it is stated in the question that 0 is required to be in the set as one of the conditions.

Jack Ceroni - 2 years, 5 months ago

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