# 2002 AIME II Problems/Problem 7

## Problem

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 $200$.

## Solution

$\frac{k(k+1)(2k+1)}{6}$ 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 $\boxed{112}$.

## Solution 2

To elaborate, we write out all 6 possibilities of parings. For example, we have

$k \equiv 24 \pmod{25}$ $k \equiv 15 \pmod{16}$

is one pairing, and

$k \equiv 24 \pmod{25}$ $k \equiv 0 \pmod{16}$

is another. We then solve this by writing the first as $16k+15 \equiv 24 \pmod{25}$ and then move the 15 to get $16k \equiv 9 \pmod{25}$.

We then list out all the mods of the multiples of $16$, and realize that each of these $6$ pairings can be generalized to become one of these multiples of $16$.

The chain is as follows:

$16 \pmod{25}$

$7, 23, 14, 5, 21, 12, 3, 19, 10, 1, 17, 8, 24, 15, 6, 22, 13, 4, 20, 11, 27, 18, 9, 0,$ and then it loops.

We see that for the first equation we have $9 \pmod {25}$ at the 24th position, so we then do $24(16)+15$ to get the first answer of 399.

Again, this is possible to repeat for all $6$ cases. CRT guarantees that we will have a solution before $25 \times 16$ or $400$ and indeed we did :P