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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7758
GMAT 1: 760 Q51 V42 GPA: 3.82
y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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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 $$y$$.

$$A. 8$$

$$B. 9$$

$$C. 10$$

$$D. 11$$

$$E. 12$$

_________________
Intern  Joined: 17 Aug 2017
Posts: 1
Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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1
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.
Manager  S
Joined: 25 Jul 2018
Posts: 203
y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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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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Intern  B
Joined: 10 Jan 2016
Posts: 4
Location: India
Schools: HEC Montreal '22
GMAT 1: 710 Q49 V37 Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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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)
Director  P
Joined: 19 Oct 2018
Posts: 791
Location: India
y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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1
$$(K-5)^2+(K-4)^2.....+(K+4)^2+(K+5)^2=y^2$$
$$11K^2+2*(1^2+2^2+3^2+4^2+5^2)=y^2$$
$$11K^2+2*(5*6*11)/6=y^2$$
$$11K^2+110=y^2$$
$$11(K^2+10)=y^2$$

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 $$y$$.

$$A. 8$$

$$B. 9$$

$$C. 10$$

$$D. 11$$

$$E. 12$$

Posted from my mobile device
Intern  B
Joined: 17 Aug 2014
Posts: 6
Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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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
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7758
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: y^2 is a perfect square which is the sum of the squares of 11 consecut  [#permalink]

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=>

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
_________________ 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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