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 $\boxed{112}$ .

[b] Addition 9/23/17 [/b] 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 mod 25 at the 21st 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*16 or 400 and indeed we did :P

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

Page

Toolbox

Search

2002 AIME II Problems/Problem 7

Problem

Solution

See also