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# What is the remainder when [m]1^2[/m] + [m]2^2[/m] + [m]3^2[/m] + [m]4

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Senior Manager
Joined: 18 Jul 2018
Posts: 484
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
What is the remainder when [m]1^2[/m] + [m]2^2[/m] + [m]3^2[/m] + [m]4  [#permalink]

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Updated on: 11 Aug 2018, 08:48
1
00:00

Difficulty:

55% (hard)

Question Stats:

36% (00:58) correct 64% (00:37) wrong based on 14 sessions

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What is the remainder when $$1^2$$ + $$2^2$$ + $$3^2$$ + $$4^2$$ + ..... + $$100^2$$ is divided by 4?

A) 0
B) 1
C) 2
D) 3
E) 4

_________________

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Originally posted by Afc0892 on 11 Aug 2018, 07:15.
Last edited by chetan2u on 11 Aug 2018, 08:48, edited 1 time in total.
Formatted question and corrected the OA
Math Expert
Joined: 02 Aug 2009
Posts: 7099
Re: What is the remainder when [m]1^2[/m] + [m]2^2[/m] + [m]3^2[/m] + [m]4  [#permalink]

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11 Aug 2018, 08:45
Afc0892 wrote:
What is the remainder when $$1^2$$ + $$2^2$$ + $$3^2$$ + $$4^2$$ + ..... + $$100^2$$ is divided by 4?

1) 0
2) 1
3) 2
4) 3

All even numbers will be multiple of 4...2^2,4^2,...100^2

Let us see the odd numbers..
$$1^2+3^2+5^2+7^2+........99^2$$
All the numbers are of the form of (2n+1)^2=4n^2+4n+1 will leave a remainder of 1
So all these 50 terms 1^2+3^2+5^2...+99^2 will leave 1+1+1..+1=1*50 as remainder

50 divided by 4 leaves a remainder of 2

C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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What is the remainder when [m]1^2[/m] + [m]2^2[/m] + [m]3^2[/m] + [m]4  [#permalink]

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11 Aug 2018, 09:32
Afc0892 wrote:
What is the remainder when $$1^2$$ + $$2^2$$ + $$3^2$$ + $$4^2$$ + ..... + $$100^2$$ is divided by 4?

A) 0
B) 1
C) 2
D) 3
E) 4

We know , Formula for the sum of the first n squares:- $$\frac{n(n+1)(2n+1)}{6}$$

$$1^2$$ + $$2^2$$ + $$3^2$$ + $$4^2$$ + ..... + $$100^2$$= $$\frac{100*101*201}{6}$$

Now, $$\frac{100*101*201}{6*4}$$= 84587*4+2

Ans. (C)
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What is the remainder when [m]1^2[/m] + [m]2^2[/m] + [m]3^2[/m] + [m]4 &nbs [#permalink] 11 Aug 2018, 09:32
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# What is the remainder when [m]1^2[/m] + [m]2^2[/m] + [m]3^2[/m] + [m]4

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