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2007 Cyprus MO/Lyceum/Problem 29

From AoPSWiki

Problem

The minimum value of a positive integer $k$ , for which the sum $\displaystyle S=k+(k+1)+(k+2)+\ldots+(k+10)$ is a perfect square, is

$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } \mathrm{Non...$

Solution

$\displaystyle k+(k+1)+(k+2)+\ldots+(k+10)=11k+55=11(k+5)$

Thus, $k+5$ must be divisible by $11$ , and the minimum for $k$ is $6\Longrightarrow\mathrm{ B}$ .

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 28	Followed by Problem 30
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 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2007_Cyprus_MO/Lyceum/Problem_29"

Category: Introductory Algebra Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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