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y^2 is a perfect square which is the sum of the squares of 11 consecut

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New post 06 Aug 2019, 01:45
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

58% (02:07) correct 42% (02:24) wrong based on 45 sessions

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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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Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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New post 06 Aug 2019, 02:16
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THIS IS POSSIBLE ONLY WHEN WE TAKE THE 11 CONSECUTIVE INTEGERS -4,-3,-2,-1,0,1,2,3,4,5,6 AND SUM OF THE SQUARES OF THEM IS 121 AND IT IS THE SQUARE OF 11. THERE FORE Y=11.
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y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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New post 06 Aug 2019, 03:21
Minimum possible value of sum of the squares of 11 consecutive numbers is equal to —> sum of the squares of : -5,-4,-3,-2,-1,0,1,2,3,4,5 —> \((-5)^{2}\)+\((-4)^{2}\)+\((-3)^{2}\)+\((-2)^{2}\)+\((-1)^{2}\)+0+\(1^{2}\)+\(2^{2}\)+\(3^{2}\)+\(4^{2}\)+\(5^{2}\)= 110

—> That means possible answer choice for y^2 should be over 110 (D or E )

The next minimum possible value:
—> \((-4)^{2}\)+\((-3)^{2}\)+\((-2)^{2}\)+\((-1)^{2}\)+0+\(1^{2}\)+\(2^{2}\)+\(3^{2}\)+\(4^{2}\)+\(5^{2}\)+\(6^{2}\)=60+61=121=\((11)^{2}\)=\((y)^{2}\)

The answer choice is D

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Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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New post 06 Aug 2019, 10:16
If I consider the consequent 11 integers as x-5, x-4,...x, x+1,...x+5 and add them, it adds up to 11x. For 11x to be a perfect square, the minimum value of x would be 11.

Hence answer would be 11(E)
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y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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New post 06 Aug 2019, 11:29
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\((K-5)^2+(K-4)^2.....+(K+4)^2+(K+5)^2=y^2\)
\(11K^2+2*(1^2+2^2+3^2+4^2+5^2)=y^2\)
\(11K^2+2*(5*6*11)/6=y^2\)
\(11K^2+110=y^2\)
\(11(K^2+10)=y^2\)

Hence y must be a multiple of 11. Minimum value that y can take is 11.

MathRevolution wrote:
[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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Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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New post 06 Aug 2019, 12:30
If we start from 1, Sum of first 11 numbers = 66 ; [(1+11)/2]
1+2+3...11 = 66
2+3+4,..12 = 77
3+4+5..13 = 88

In above series, the first perfect square is 121.. Which is a square of 11..

So my answer is D
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Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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New post 08 Aug 2019, 00:25
=>

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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Re: y^2 is a perfect square which is the sum of the squares of 11 consecut   [#permalink] 08 Aug 2019, 00:25
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