Problem:
Find the number of permutations of numbers such that the sum of five products
is divisible by .
Solution:
Answer (080):
Because the given condition is cyclic, it is sufficient to count the number of permutations satisfying the condition with and then multiply the result by 5 . Under this additional condition,
Because neither nor is a multiple of 3 , it is necessary and sufficient for to be a multiple of 3 . From the remaining numbers, there are two that are congruent to and two that are congruent to . Hence there are 4 choices for choices for given choices for , leaving one choice for for a total of choices.
Hence there are permutations satisfying the given condition.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions