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

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== Solution ==
 
== Solution ==
Note that <math>k^2_1 + k^2_2 + ... + k^2_n \leq \frac{k(k+1)(2k+1)}{6}</math>.
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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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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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When <math>n = 18</math>, <math>\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002</math>.
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Therefore, we know <math>n leq 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 20:20, 10 March 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 $n leq 17$.

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
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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