2002 AMC 12P Problems/Problem 13: Difference between revisions
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== Solution == | == Solution == | ||
Note that | Note that <math>k^2_1 + k^2_2 + ... + k^2_n <= \frac{k(k+1)(2k+1)}{6}</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 19:16, 10 March 2024
Problem
What is the maximum value of 
for which there is a set of distinct positive integers 
for which
![\[k^2_1 + k^2_2 + ... + k^2_n = 2002?\]](http://latex.artofproblemsolving.com/7/d/a/7da913f9b8644e18078f96cf0ab0c4e928633633.png)
Solution
Note that 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
