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2002 AMC 12P Problems/Problem 13

Revision as of 02:03, 2 July 2024 by The 76923th (talk | contribs) (→‎Solution)

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$ (Error compiling LaTeX. Unknown error_msg)a > 17$to account for the amount under$2002$, which is$2002-1785 = 217$.

Essentially, this boils down to writing$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)(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:$ (Error compiling LaTeX. Unknown error_msg)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:$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)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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