2002 AIME II Problems/Problem 7
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Problem
It is known that, for all positive integers ,
![$1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6$](http://latex.artofproblemsolving.com/4/f/c/4fc69b2d25fe9c2e53533b462d9b0cce6f4ceed0.png)
Find the smallest positive integer such that
is a multiple of
.
Solution
is a multiple of
if
is a multiple of
.
So
.
Since is always odd, and only one of
and
is even, either
.
Thus, .
If , then
. If
, then
. If
, then
.
Thus, there are no restrictions on in
.
Ii is easy to see that only one of ,
, and
is divisible by
. So either
.
Thus, .
From the Chinese Remainder Theorem, . Thus, the smallest positive integer
is
.
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |