A group of children held a grape-eating contest.
When the contest was over, the winner had eaten N grapes, where N is a positive integer.
The child in Mth place had eaten N + 2 – 2M grapes.
The total number of grapes eaten in the contest was 2016.
Find the smallest possible value of N.
I am solving this problem with the provision that each child ate a different number of grapes.
The number of grapes eaten by each child's place forms an arithmetic sequence: N, N-2, N-4, ... N-2(M-1), assuming there are M children. The sum of the sequence is M*(N + N-2*(M-1))/2 = M*(N-M+1).
Then M*(N-M+1) = 2016 implies N = 2016/M + M - 1. For N to be an integer, M must be a factor of 2016. Subsequently the sum 2016/M + M is the sum of a two term factorization of 2016.
In a continuous system the minimum value occurs when M = sqrt(2016) = 44.899889. In the integers, the closest we can get is 2016 = 42*48. Then N = 42 + 48 - 1 = 89.
This checks out: with 42 children and the first place child eating 89 grapes, the last place child eats 7 grapes and the sum of the series 89+87+...+7 = 2016.
Solution with no ties
