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If m and n are positive integers and m^2 + n^2 = 40, what is

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If m and n are positive integers and m^2 + n^2 = 40, what is [#permalink]

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New post 01 Nov 2012, 06:56
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If m and n are positive integers and m^2 + n^2 = 40, what is the value of m^3 + n^3?

A. 72
B. 224
C. 320
D. 512
E. 1,600

Pls try algebra method...
Of course, plugging nos easily gives 2 and 6(m and n being integers), but i attempted algebra first
and was wasting time.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 01 Nov 2012, 07:01, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If m and n are positive integers and m^2 + n^2 = 40, what is [#permalink]

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New post 01 Nov 2012, 11:16
Hey,

i have tried it this way:

You need to integers which squared are equal 40.

Which could it be ? Let's start with the first integer:

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
Stop. The integers can't be greater than 6 or we will score above 40.

The second integer need to be picked up the same way.
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36

The only pair that matches is 6^2 + 2^2 = 40.

So 6^3 + 2^3 = 224.

Answer b.)
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Re: If m and n are positive integers and m^2 + n^2 = 40, what is [#permalink]

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New post 10 Dec 2012, 22:35
I tried algebraic method. I think it will eat up my life in the test. Haha! Nonetheless, I would like to see how that is done.

I just selected from the possible values of \(m^2\) and \(n^2\): \(1,4,9,16,25,36\) and selected \(2^2\) and \(6^2\)

\(m^3 + n^3=2^3 + 6^3 = 224\)
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Re: If m and n are positive integers and m^2 + n^2 = 40, what is [#permalink]

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New post 11 Aug 2013, 03:57
If you try to solve it through algebra you will definitely spend the whole 75 minutes on it alone :-D ..while picking numbers will help you to solve easily..just see which numbers fit as M and N to make the given equation possible...
Re: If m and n are positive integers and m^2 + n^2 = 40, what is   [#permalink] 11 Aug 2013, 03:57
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