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

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Director
Joined: 25 Oct 2008
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If m and n are both positive integers and m>n is 6 a [#permalink]

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18 Jul 2009, 01:32
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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?
_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Kudos [?]: 1182 [0], given: 100

Manager
Joined: 07 Apr 2009
Posts: 145

Kudos [?]: 13 [0], given: 3

Re: m and n are both positive integers [#permalink]

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18 Jul 2009, 08:53
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

Kudos [?]: 13 [0], given: 3

Manager
Joined: 03 Jul 2009
Posts: 104

Kudos [?]: 94 [0], given: 13

Location: Brazil
Re: m and n are both positive integers [#permalink]

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18 Jul 2009, 09:26
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.

Kudos [?]: 94 [0], given: 13

Re: m and n are both positive integers   [#permalink] 18 Jul 2009, 09:26
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# If m and n are both positive integers and m>n is 6 a

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