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

Revision as of 18:33, 23 September 2017

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

@@ Line 21: / Line 21: @@
 From the [[Chinese Remainder Theorem]], <math>k \equiv 0, 112, 224, 175, 287, 399 \pmod{400}</math>. Thus, the smallest positive integer <math>k</math> is <math>\boxed{112}</math>.
+[b] Addition 9/23/17 [/b]
+To elaborate, we write out all 6 possibilities of parings. For example, we have
+<math>k \equiv 24 \pmod{25}</math>
+<math>k \equiv 15 \pmod(16)</math>
+is one pairing, and
+<math>k \equiv 24 \pmod{25}</math>
+<math>k \equiv 0 \pmod{16}</math>
+is another. We then solve this by writing the first as <math>16k+15 \equiv 24 \pmod{25}</math> and then move the 15 to get <math>16k \equiv 9 \pmod{25}</math>.
+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:
+<math> 16 \pmod{25}</math>
+, 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 <math>24(16)+15</math> 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
 == See also ==
 {{AIME box|year=2002|n=II|num-b=6|num-a=8}}
 {{MAA Notice}}

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

Revision as of 18:33, 23 September 2017

Problem

Solution

See also