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y^2 is a perfect square which is the sum of the squares of 1

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y^2 is a perfect square which is the sum of the squares of 1

by Max@Math Revolution » Tue Aug 06, 2019 12:32 am

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[GMAT math practice question]

y^2 is a perfect square which is the sum of the squares of 11 consecutive numbers. Find the minimum possible value for the positive integer y.

A. 8
B. 9
C. 10
D. 11
E. 12
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by Max@Math Revolution » Wed Aug 07, 2019 11:06 pm
=>

Let the consecutive numbers be n-5, n-4, ..., n, ..., n+4, and n+5. Then
y^2 = (n-5)^2 + (n-4)^2 + (n-3)^2 + (n-2)^2 + (n-1)^2 + n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 + (n+5)^2
= n^2 -10n + 25 + n^2 - 8n + 16 + n^2 -6n + 9 + n^2 - 4n + 4 + n^2 - 2n + 1 + n^2 + n^2 +2n + 1 + n^2 +4n + 4 + n^2 +6n + 9 + n^2 +8n + 16 + n^2 +10n + 25
= 11n^2 + 110

If n = 0, then 11*0^2 + 110 = 110 is not a perfect square.
If n = 1, then 11*1^2 + 110 = 121 is the perfect square of 11.
Thus, y = 11.

Therefore, D is the answer.
Answer: D
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