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

Revision as of 20:20, 10 March 2024

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

$k^2_1 + k^2_2 + ... + k^2_n = 2002?$

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

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 $n leq 17$ .

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 17: / Line 17: @@
 == Solution ==
-Note that <math>k^2_1 + k^2_2 + ... + k^2_n \leq \frac{k(k+1)(2k+1)}{6}</math>.
+Note that <math>k^2_1 + k^2_2 + ... + k^2_n = 2002 \leq \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>.
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=12|num-a=14}}
 {{MAA Notice}}