Problem:
Find the largest possible value of k for which 311 is expressible as the sum of k consecutive positive integers.
Solution:
To solve the problem, we must find integers n and k such that n is nonnegative, k is as large as possible, and
311=(n+1)+(n+2)+⋯+(n+k)(1)
Noting that
(n+1)+(n+2)+⋯+(n+k)=[1+2+⋯+(n+k)]−[1+2+⋯+n]=2(n+k)(n+k+1)−2n(n+1)=2k(k+2n+1)
it follows that (1) is equivalent to
k(k+2n+1)=2⋅311(2)
In solving (2), we must ensure that the smaller factor, k, is as large as possible, and that n is a non-negative integer. These conditions lead to k=2⋅35=486,n=121 and 311=122+123+⋯+607.
OR
Let m be the average of the k consecutive integers. If k is odd, then m must be the middle integer, and km=311. Now k=35 and m=36 is the best we can do, for if k=36 then m−(k−1)/2, the smallest summand, is negative. But if k is even, then m lies halfway between the middle two integers in the sum. Thus (2m)k=2⋅311 and now the largest even divisor of 2⋅311 which does not give rise to a negative first summand is 2⋅35=486. This is the answer.
The problems on this page are the property of the MAA's American Mathematics Competitions