2002 AMC 12P Problems/Problem 13: Difference between revisions

Latest revision as of 20:48, 15 October 2025

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 $\boxed {\textbf{(D) }17}$ .

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #13]] and [[2002 AMC 10P Problems|2002 AMC 10P #24]]}}
 == Problem ==
-How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
+What is the maximum value of <math>n</math> for which there is a set of distinct positive integers <math>k_1, k_2, ... k_n</math> for which
+<cmath>k^2_1 + k^2_2 + ... + k^2_n = 2002?</cmath>
-<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }  </math>
+<math>
+\text{(A) }14
+\qquad
+\text{(B) }15
+\qquad
+\text{(C) }16
+\qquad
+\text{(D) }17
+\qquad
+\text{(E) }18
+</math>
 == Solution ==
-If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+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 = 18</math>, <math>\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002</math>.
+Therefore, we know <math>n \leq 17</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>.
+We can rewrite this as <math>(a+b)(a-b) = 217</math>. Since <math>217 = 7 \cdot 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>
+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>
+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>. Since <math>n \geq 18</math> is impossible, the answer is <math>\boxed {\textbf{(D) }17}</math>.
 == See also ==
-{{AMC12 box|year=2000|num-b=6|num-a=8}}
+{{AMC10 box|year=2002|ab=P|num-b=23|num-a=25}}
+{{AMC12 box|year=2002|ab=P|num-b=12|num-a=14}}
 {{MAA Notice}}
+[[Category: Introductory Algebra Problems]]

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2002 AMC 12P Problems/Problem 13: Difference between revisions

Latest revision as of 20:48, 15 October 2025

Problem

Solution

See also