Join us for MBA Spotlight – The Top 20 MBA Fair      Schedule of Events | Register

 It is currently 05 Jun 2020, 11:49

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 9032
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]

Show Tags

06 Aug 2019, 00:45
00:00

Difficulty:

65% (hard)

Question Stats:

57% (02:19) correct 43% (02:26) wrong based on 53 sessions

HideShow timer Statistics

[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 $$y$$.

$$A. 8$$

$$B. 9$$

$$C. 10$$

$$D. 11$$

$$E. 12$$

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" 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] Show Tags 06 Aug 2019, 01:16 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. Director Joined: 25 Jul 2018 Posts: 719 y^2 is a perfect square which is the sum of the squares of 11 consecut [#permalink] Show Tags 06 Aug 2019, 02:21 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 Posted from my mobile device Intern Joined: 10 Jan 2016 Posts: 4 Location: India GMAT 1: 710 Q49 V37 Re: y^2 is a perfect square which is the sum of the squares of 11 consecut [#permalink] Show Tags 06 Aug 2019, 09:16 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) DS Forum Moderator Joined: 19 Oct 2018 Posts: 1888 Location: India y^2 is a perfect square which is the sum of the squares of 11 consecut [#permalink] Show Tags 06 Aug 2019, 10:29 1 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 Joined: 17 Aug 2014 Posts: 6 Location: India Schools: Cass '21 (A) GMAT 1: 680 Q47 V35 GPA: 3.83 Re: y^2 is a perfect square which is the sum of the squares of 11 consecut [#permalink] Show Tags 06 Aug 2019, 11:30 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 9032 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] Show Tags 07 Aug 2019, 23:25 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Re: y^2 is a perfect square which is the sum of the squares of 11 consecut   [#permalink] 07 Aug 2019, 23:25