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Difference between revisions of "2002 AMC 12P Problems/Problem 13"

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Therefore, we know <math>n \leq 17</math>.
 
Therefore, we know <math>n \leq 17</math>.
  
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Now we must show that <math>n = 17</math> works. We replace some integer <math>b</math> within the set <math>\{1, 2, ... 17}\$ with an integer </math>a > 17<math> to account for the amount under </math>2002<math>, which is </math>2002-1785 = 217<math>.
+
Now we must show that <math>n = 17</math> works. We replace some integer <math>b</math> within the set <math>{1, 2, ... 17}</math> with an integer <math>a > 17</math> to account for the amount under <math>2002</math>, which is <math>2002-1785 = 217</math>.
  
−
Essentially, this boils down to writing </math>217<math> as a difference of squares. Assume there exist positive integers </math>a<math> and </math>b<math> where </math>a > 17<math> and </math>b \leq 17<math> such that </math>a^2 - b^2 = 217<math>.
+
Essentially, this boils down to writing <math>217</math> as a difference of squares. Assume there exist positive integers <math>a</math> and <math>b</math> where <math>a > 17</math> and <math>b \leq 17</math> such that <math>a^2 - b^2 = 217</math>.
  
−
We can rewrite this as </math>(a+b)(a-b) = 217<math>. Since </math>217 = (7)(31)<math>, either </math>a+b = 217<math> and </math>a-b = 1<math> or </math>a+b = 31<math> and </math>a-b = 7<math>. We analyze each case separately.
+
We can rewrite this as <math>(a+b)(a-b) = 217</math>. Since <math>217 = (7)(31)</math>, either <math>a+b = 217</math> and <math>a-b = 1</math> or <math>a+b = 31</math> and <math>a-b = 7</math>. We analyze each case separately.
  
−
Case 1: </math>a+b = 217<math> and </math>a-b = 1<math>
+
Case 1: <math>a+b = 217</math> and <math>a-b = 1</math>
  
−
Solving this system of equations gives </math>a = 109<math> and </math>b = 108<math>. However, </math>108 > 17<math>, so this case does not yield a solution.
+
Solving this system of equations gives <math>a = 109</math> and <math>b = 108</math>. However, <math>108 > 17</math>, so this case does not yield a solution.
  
−
Case 2: </math>a+b = 31<math> and </math>a-b = 7<math>
+
Case 2: <math>a+b = 31</math> and <math>a-b = 7</math>
  
−
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.
  
−
The list </math>1, 2 ... 11, 13, 14 ... 17, 19<math> has </math>17<math> terms whose sum of squares equals </math>2002<math>. Therefore, the answer is </math>\boxed {\text{(D) }17}$.
+
The list <math>1, 2 ... 11, 13, 14 ... 17, 19</math> has <math>17</math> terms whose sum of squares equals <math>2002</math>. Therefore, the answer is <math>\boxed {\text{(D) }17}</math>.
  
 
== See also ==
 
== See also ==
 
{{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}}

Revision as of 02:04, 2 July 2024

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

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 $217 = (7)(31)$, 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 $2002$. Therefore, the answer is $\boxed {\text{(D) }17}$.

See also

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

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