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

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== Problem ==
 
== Problem ==
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It is known that, for all positive integers <math>k</math>,
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<center><math>1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6</math>.</center>
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Find the smallest positive integer <math>k</math> such that <math>1^2+2^2+3^2+\ldots+k^2</math> is a multiple of <math>200</math>.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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<math>\frac{k(k+1)(2k+1)}{6}</math> is a multiple of <math>200</math> if <math>k(k+1)(2k+1)</math> is a multiple of <math>1200 = 2^4 \cdot 3 \cdot 5^2</math>.
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So <math>16,3,25|k(k+1)(2k+1)</math>.
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Since <math>2k+1</math> is always odd, and only one of <math>k</math> and <math>k+1</math> is even, either <math>k, k+1 \equiv 0 \pmod{16}</math>.
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Thus, <math>k \equiv 0, 15 \pmod{16}</math>.
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If <math>k \equiv 0 \pmod{3}</math>, then <math>3|k</math>. If <math>k \equiv 1 \pmod{3}</math>, then <math>3|2k+1</math>. If <math>k \equiv 2 \pmod{3}</math>, then <math>3|k+1</math>.
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Thus, there are no restrictions on <math>k</math> in <math>\pmod{3}</math>.
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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>.
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Thus, <math>k \equiv 0, 24, 12 \pmod{25}</math>.
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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>.
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== Solution 2==
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To elaborate, we write out all 6 possibilities of pairings. For example, we have
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<math>k \equiv 24 \pmod{25}</math>
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<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 <math>16</math>, and realize that each of these <math>6</math> pairings can be generalized to become one of these multiples of <math>16</math>.
 +
 
 +
The chain is as follows:
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 +
<math>16 \pmod{25}</math>
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<math>7, 23, 14, 5, 21, 12, 3, 19, 10, 1, 17, 8, 24, 15, 6, 22, 13, 4, 20, 11, 27, 18, 9, 0,</math> and then it loops.
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We see that for the first equation we have <math>9 \pmod {25}</math> at the 24th position, so we then do <math>24(16)+15</math> to get the first answer of 399.
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Again, this is possible to repeat for all <math>6</math> cases. [[Chinese Remainder Theorem|CRT]] guarantees that we will have a solution before <math>25 \times 16</math> or <math>400</math> and indeed we did :P
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== See also ==
 
== See also ==
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* [[2002 AIME II Problems/Problem 6 | Previous problem]]
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{{AIME box|year=2002|n=II|num-b=6|num-a=8}}
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* [[2002 AIME II Problems/Problem 8 | Next problem]]
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* [[2002 AIME II Problems]]
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[[Category: Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 11:10, 9 September 2023

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 pairings. 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

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

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