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 is equal to kn. 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.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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