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# What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ?

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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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Updated on: 09 Apr 2018, 07:08
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What is the value of $$38^2 + 39^2 + 40^2 + 41^2 + 42^2$$ ?

A. 7950
B. 7990
C. 8010
D. 8050
E. 8070
[Reveal] Spoiler: OA

Originally posted by selim on 09 Apr 2018, 06:55.
Last edited by Bunuel on 09 Apr 2018, 07:08, edited 1 time in total.
Renamed the topic and edited the question.
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Posts: 44635
Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 07:06
selim wrote:
what is the value of 38^2 + 39^2+40^2 +41^2 + 42^2

A.7950
B.7990
C.8010
D.8050
E.8070

This is a copy of this question: https://gmatclub.com/forum/36-126078.html and this one: https://gmatclub.com/forum/what-is-the- ... 98546.html
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Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 07:48
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selim wrote:
What is the value of $$38^2 + 39^2 + 40^2 + 41^2 + 42^2$$ ?

A. 7950
B. 7990
C. 8010
D. 8050
E. 8070

I love this question (and all of the replicas )!!

First notice that 38² + 39² + 40² + 41² + 42² = (40 - 2)² + (40 - 1)² + 40² + (40 + 1)² + (40 + 2)²

Now notice that (40 - 2)² + (40 - 1)² + 40² + (40 + 1)² + (40 + 2)² has the form (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)²
So, let's expand and simplify (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² and see what we get.

(x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² = [x² - 4x + 4] + [x² - 2x + 1] + [x²] + [x² + 2x + 1] + [x² + 4x + 4] [lots of canceling to do here...]
= 5x² + 10

So, if (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² simplifies to become 5x² + 10...
...then (40 - 2)² + (40 - 1)² + 40² + (40 + 1)² + (40 + 2)² must simplify to become 5(40²) + 10

5(40²) + 10 = 5(1600) + 10
= 8000 + 10
= 8010

Cheers,
Brent
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Posts: 5777
What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 08:43
selim wrote:
What is the value of $$38^2 + 39^2 + 40^2 + 41^2 + 42^2$$ ?

A. 7950
B. 7990
C. 8010
D. 8050
E. 8070

It is always good to know the formula for SUM of square of first n natural numbers.. $$\frac{n(n+1)(2n+1)}{6}$$..

so $$38^2 + 39^2 + 40^2 + 41^2 + 42^2$$ = ($$1^2+2^2+............. 41^2 + 42^2$$)-($$1^2 + 2^2 + 3^2 +...........+ 36^2 + 37^2$$)..

$$\frac{42(42+1)(2*42+1)}{6}-\frac{37(37+1)(2*37+1)}{6}=7*43*85-37*19*25=8010$$
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 09:02
My method:

(30+8) + (30+9) + (30+10) + (30+11) + 30+12) . I just did 8^2+9^2+10^2+11^2+12^2 = ...10, so the number should end with 10. The answer is C.

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Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 09:20
I don't think this is right approach, this approach could have worked if we want to eliminate some options which do
not end with 0. here all the options end with 0. so this approach might not work.

Experts view needed Bunuel, @chetan4u

Iamnowjust wrote:
My method:

(30+8) + (30+9) + (30+10) + (30+11) + 30+12) . I just did 8^2+9^2+10^2+11^2+12^2 = ...10, so the number should end with 10. The answer is C.

Posted from my mobile device

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Senior Manager
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WE: Business Development (Energy and Utilities)
Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 09:24
I really like your approach using this formula.
In fact I want to add 2 more fundamental formula for summation:

1) $$1+2+3+...+n = n(n+1)/2$$ - Sum of first n natural numbers.
2) $$1^3+2^3+...+n^3 = (n(n+1)/2)^2$$ - Sum of cube of first n natural numbers.

chetan2u wrote:
selim wrote:
What is the value of $$38^2 + 39^2 + 40^2 + 41^2 + 42^2$$ ?

A. 7950
B. 7990
C. 8010
D. 8050
E. 8070

It is always good to know the formula for SUM of square of first n natural numbers.. $$\frac{n(n+1)(2n+1)}{6}$$..

so $$38^2 + 39^2 + 40^2 + 41^2 + 42^2$$ = $$1^2+2^2+.......38^2 + 39^2 + 40^2 + 41^2 + 42^2$$-($$1^2 + 2^2 + 3^2 +...........+ 36^2 + 37^2$$..

$$\frac{42(42+1)(2*42+1)}{6}-\frac{37(37+1)(2*37+1)}{6}=7*43*85-37*19*25=8010$$

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Posts: 5777
Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 09:45
Iamnowjust wrote:
My method:

(30+8) + (30+9) + (30+10) + (30+11) + 30+12) . I just did 8^2+9^2+10^2+11^2+12^2 = ...10, so the number should end with 10. The answer is C.

Posted from my mobile device

You are lucky to have got the correct answer..

There is no logic behind this..
example 38^2+39^2=2965, ...so __65
8^2+9^2=145

so you can just talk of UNITS digit and NOT tens digit
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Joined: 08 Sep 2016
Posts: 62
Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? [#permalink]

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09 Apr 2018, 10:28
(40-2)^2 + (40-1)^2 + 40^2 + (40+1)^2 + (40+2)^2

You can foil each expression for the answer. A quick way to the answer is to focus on the last multiplication when you foil for each expression.

You will get 4+1+4+1 = 10

Check out the link that Bunuel provided. His/her approach to this problem is really good.
https://gmatclub.com/forum/36-126078.html
Re: What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ?   [#permalink] 09 Apr 2018, 10:28
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