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If m and n are both positive integers and m>n is 6 a

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Director
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If m and n are both positive integers and m>n is 6 a [#permalink] New post 18 Jul 2009, 00:32
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If m and n are both positive integers and m>n is 6 a factor of the product mn?
1. m+n=188
2. m is 150% of n

[Reveal] Spoiler:
Ans:B


How is b sufficient guys?
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Re: m and n are both positive integers [#permalink] New post 18 Jul 2009, 07:53
I will try to explain..

for option B, when m = 150/100*n ==>2m = 3n

this will be valid only when m = 3 & n = 2

and so mn = 6 is always divisble by 6
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Re: m and n are both positive integers [#permalink] New post 18 Jul 2009, 08:26
Well, actually is because m is a multiple of 3, an n is a multiple of 2.
m = 3, 6, 9, 12 ....
n = 2, 4, 6, 8 ...

Step by step.

If m = 150% of n,
m = 150/100 * n = 3/2 * n

If 6 is a multiple of m*n, then mn = 6k, k an integer.
Substituting m for n you have

3/2 * n * n = 6k
Resolving for n you have n = 2 \sqrt{k}

Because m = 3/2 *n
m = 3 * \sqrt{k}

As k, m and n are integers, \sqrt{k} must be an integer as well.
So m is a multiple of 3, and n is a multiple of 2.

If you multiple a multiple of 2 with a multiple of 3, you will have a multiple of 6, thus mn is a multiple of 6 and b is sufficient.

I hope to have helped you.
Re: m and n are both positive integers   [#permalink] 18 Jul 2009, 08:26
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If m and n are both positive integers and m>n is 6 a

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