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Difference between revisions of "2002 AMC 12P Problems/Problem 13"
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== Solution ==
 
== Solution ==
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Note that <math>k^2_1 + k^2_2 + ... + k^2_n = 2002 \leq \frac{n(n+1)(2n+1)}{6}</math>
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Note that <math>k^2_1 + k^2_2 + ... + k^2_n = 2002 \geq \frac{n(n+1)(2n+1)}{6}</math>
  
 
When <math>n = 17</math>, <math>\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002</math>.
 
When <math>n = 17</math>, <math>\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002</math>.
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Solving this system of equations gives <math>a = 19</math> and <math>b = 12</math>. This satisfies all the requirements of the problem.
 
Solving this system of equations gives <math>a = 19</math> and <math>b = 12</math>. This satisfies all the requirements of the problem.
  
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The list <math>1, 2 ... 11, 13, 14 ... 17, 19</math> has <math>17</math> terms whose sum of squares equals <math>2002</math>. Since <math>n \geq 18</math> is impossible, the answer is <math>\boxed {\text{(D) }17}</math>.
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The list <math>1, 2 ... 11, 13, 14 ... 17, 19</math> has <math>17</math> terms whose sum of squares equals <math>2002</math>. Since <math>n \geq 18</math> is impossible, the answer is <math>\boxed {\textbf{(D) }17}</math>.
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2002|ab=P|num-b=23|num-a=25}}
 
{{AMC12 box|year=2002|ab=P|num-b=12|num-a=14}}
 
{{AMC12 box|year=2002|ab=P|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:20, 12 October 2024

The following problem is from both the 2002 AMC 12P #13 and 2002 AMC 10P #24, so both problems redirect to this page.

Problem

What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1, k_2, ... k_n$ for which

\[k^2_1 + k^2_2 + ... + k^2_n = 2002?\]

$\text{(A) }14 \qquad \text{(B) }15 \qquad \text{(C) }16 \qquad \text{(D) }17 \qquad \text{(E) }18$

Solution

Note that $k^2_1 + k^2_2 + ... + k^2_n = 2002 \geq \frac{n(n+1)(2n+1)}{6}$

When $n = 17$, $\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002$.

When $n = 18$, $\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002$.

Therefore, we know $n \leq 17$.

Now we must show that $n = 17$ works. We replace some integer $b$ within the set $\{1, 2, ... 17\}$ with an integer $a > 17$ to account for the amount under $2002$, which is $2002-1785 = 217$.

Essentially, this boils down to writing $217$ as a difference of squares. Assume there exist positive integers $a$ and $b$ where $a > 17$ and $b \leq 17$ such that $a^2 - b^2 = 217$.

We can rewrite this as $(a+b)(a-b) = 217$. Since $217 = 7 \cdot 31$, either $a+b = 217$ and $a-b = 1$ or $a+b = 31$ and $a-b = 7$. We analyze each case separately.

Case 1: $a+b = 217$ and $a-b = 1$

Solving this system of equations gives $a = 109$ and $b = 108$. However, $108 > 17$, so this case does not yield a solution.

Case 2: $a+b = 31$ and $a-b = 7$

Solving this system of equations gives $a = 19$ and $b = 12$. This satisfies all the requirements of the problem.

The list $1, 2 ... 11, 13, 14 ... 17, 19$ has $17$ terms whose sum of squares equals $2002$. Since $n \geq 18$ is impossible, the answer is $\boxed {\textbf{(D) }17}$.

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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