1987 AIME Problems/Problem 11
Find the largest possible value of for which is expressible as the sum of consecutive positive integers.
Let us write down one such sum, with terms and first term :
Thus so is a divisor of . However, because we have so . Thus, we are looking for large factors of which are less than . The largest such factor is clearly ; for this value of we do indeed have the valid expression , for which .
First note that if is odd, and is the middle term, the sum is equal to kn. If is even, then we have the sum equal to which is going to be even. Since is odd, we see that is odd.
Thus, we have . Also, note Subsituting , we have . Proceed as in solution 1.
|1987 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.