1987 AIME Problems/Problem 11
Problem
Find the largest possible value of for which is expressible as the sum of consecutive positive integers.
Solutions
Solution 1
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 .
Solution 2
First note that if is odd, and is the middle term, the sum equals . If is even, then we have the sum equal to , which will be even. Since is odd, we see that is odd.
Thus, we have . Also, note Subsituting , we have . Proceed as in solution 1.
Solution 3
Proceed as in Solution 1 until it is noted that is a divisor of . The divisors of are . Note that the factors of are of opposite parity (if is odd, then is even and vice versa). Thus, one of the two factors will be a power of three, and the other will be twice a power of three. will represent the greater factor while will represent the lesser factor. Given this information, we need to find the factor pair that maximizes the lesser of the two factors, as this will maximize the value of . The factor pair which maximizes the lesser factor is and . It follows that = = .
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See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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