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Given that 1+2+.......+n=n(n+1)/2 and

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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26 Jan 2018, 01:21
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35% (medium)

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72% (02:00) correct 28% (02:14) wrong based on 57 sessions

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[GMAT math practice question]

Given that $$1+2+.......+n=\frac{n(n+1)}{2}$$ and $$1^2+2^2+…..+n^2=\frac{n(n+1)(2n+1)}{6}$$, what is the sum of the integers between $$1$$ and $$100$$ (inclusive) that are not squares of integers?

$$A. 4050$$
$$B. 4665$$
$$C. 4775$$
$$D. 5000$$
$$E. 5050$$

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 17 Oct 2016 Posts: 325 Location: India Concentration: Operations, Strategy GPA: 3.73 WE: Design (Real Estate) Re: Given that 1+2+.......+n=n(n+1)/2 and [#permalink] Show Tags 26 Jan 2018, 01:26 B Sum of integer without the sum of the square of the numbers = sum of numbers up to 100 -sum of square of the numbers up to 10. Hence the answer is 100(101)/2-10(11)(21)/6 =4665 Posted from my mobile device _________________ Help with kudos if u found the post useful. Thanks BSchool Forum Moderator Joined: 26 Feb 2016 Posts: 3030 Location: India GPA: 3.12 Given that 1+2+.......+n=n(n+1)/2 and [#permalink] Show Tags 26 Jan 2018, 04:06 MathRevolution wrote: [GMAT math practice question] Given that $$1+2+.......+n=\frac{n(n+1)}{2}$$ and $$1^2+2^2+…..+n^2=\frac{n(n+1)(2n+1)}{6}$$, what is the sum of the integers between $$1$$ and $$100$$ (inclusive) that are not squares of integers? $$A. 4050$$ $$B. 4665$$ $$C. 4775$$ $$D. 5000$$ $$E. 5050$$ The numbers between 1 and 100 which are squares are 1,4,9,16,25,36,49,64,81,100 The sum of these numbers is $$10 + 20 + 110 + 245 = 385$$ From the formula above, we can find the sum of all integers - $$\frac{100*101}{2} = 50*101 = 5050$$ Hence, the sum of integers between 1 and 100 that are not squares of integers are 5050 - 385 = 4665(Option B) _________________ You've got what it takes, but it will take everything you've got Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6006 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Given that 1+2+.......+n=n(n+1)/2 and [#permalink] Show Tags 28 Jan 2018, 18:47 => Since there are 10 squares of integers between $$1$$ and $$100,$$ inclusive, we need to find $$1+2+…+100 – (1^2+^22+…+10^2) = \frac{(100*101)}{2} – \frac{(10*11*21)}{6} = 5050 – 385 = 4665$$ Therefore, the answer is B. Answer : B _________________ 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$99 for 3 month Online Course"
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Re: Given that 1+2+.......+n=n(n+1)/2 and &nbs [#permalink] 28 Jan 2018, 18:47
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