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## mn problem

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mn problem Fri May 14, 2010 9:45 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
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

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Sat May 15, 2010 12:07 pm
Examine (1); 1 + 187 = 188

187 * 1 = 187

187 / 6 = 31 + 1/6

But,

186 + 2 =188, so

186 * 2 = 372, and

372 / 6 = 62

So, (1) is insufficient.

(2) say that m = 1.5n, so

Because we know that m and n must b integers we can say that the very minimum m and n could be is,

m=3, n=2, 3=1.5*2

The astute observer would also note at this point that any other combination of m and n will be multiple of m and n. We can say that the ratio is

3m:2n

More importantly we can see,

3m*2n=6mn

Thus any combination of m and n is going to result in a multiple of six (because we're multiplying any combination by 6. Thus (2) is sufficient.

Could you please confirm that B is the OA?

Hope this helps.

Thanks,

Jared

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Tue May 01, 2012 5:58 pm
6 is a factor of product m*n -> m or n at least has 2
1. If m, n is odd -> no 2 factor -> not sufficient
2. m = 3/2 n -> 2m = 3n. n absolutely has 2 as its factor and m has 3 as its factor -> n=2k and m=3g -> m*n = 6*k*g, k,g in N.

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