Difference between revisions of "2002 AIME II Problems/Problem 7"

Revision as of 00:34, 27 December 2014

It is known that, for all positive integers $k$ ,

$1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6$ .

Find the smallest positive integer $k$ such that $1^2+2^2+3^2+\ldots+k^2$ is a multiple of $200$ .

$\frac{k(k+1)(2k+1)}{6}$ is a multiple of $200$ if $k(k+1)(2k+1)$ is a multiple of $1200 = 2^4 \cdot 3 \cdot 5^2$ . So $16,3,25|k(k+1)(2k+1)$ .

Since $2k+1$ is always odd, and only one of $k$ and $k+1$ is even, either $k, k+1 \equiv 0 \pmod{16}$ .

Thus, $k \equiv 0, 15 \pmod{16}$ .

If $k \equiv 0 \pmod{3}$ , then $3|k$ . If $k \equiv 1 \pmod{3}$ , then $3|2k+1$ . If $k \equiv 2 \pmod{3}$ , then $3|k+1$ .

Thus, there are no restrictions on $k$ in $\pmod{3}$ .

It is easy to see that only one of $k$ , $k+1$ , and $2k+1$ is divisible by $5$ . So either $k, k+1, 2k+1 \equiv 0 \pmod{25}$ .

Thus, $k \equiv 0, 24, 12 \pmod{25}$ .

From the Chinese Remainder Theorem, $k \equiv 0, 112, 224, 175, 287, 399 \pmod{400}$ . Thus, the smallest positive integer $k$ is $\boxed{112}$ .

@@ Line 16: / Line 16: @@
 Thus, there are no restrictions on <math>k</math> in <math>\pmod{3}</math>.
-Ii is easy to see that only one of <math>k</math>, <math>k+1</math>, and <math>2k+1</math> is divisible by <math>5</math>. So either <math>k, k+1, 2k+1 \equiv 0 \pmod{25}</math>.
+It is easy to see that only one of <math>k</math>, <math>k+1</math>, and <math>2k+1</math> is divisible by <math>5</math>. So either <math>k, k+1, 2k+1 \equiv 0 \pmod{25}</math>.
 Thus, <math>k \equiv 0, 24, 12 \pmod{25}</math>.

2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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