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

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Math Revolution GMAT Instructor
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26 Jan 2018, 00:21
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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?

$$A. 4050$$
$$B. 4665$$
$$C. 4775$$
$$D. 5000$$
$$E. 5050$$
[Reveal] Spoiler: OA

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Re: Given that 1+2+.......+n=n(n+1)/2 and [#permalink]

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26 Jan 2018, 00:26
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

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26 Jan 2018, 03:06
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)
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Re: Given that 1+2+.......+n=n(n+1)/2 and [#permalink]

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28 Jan 2018, 17:47
=>

Since there are 10 squares of integers between $$1$$ and $$100,$$ inclusive, we need to find

$$1+2+…+100 – (1^2+^22+…+10^2) = \frac{(100*101)}{2} – \frac{(10*11*21)}{6} = 5050 – 385 = 4665$$

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Re: Given that 1+2+.......+n=n(n+1)/2 and   [#permalink] 28 Jan 2018, 17:47
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