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Source: — Data Sufficiency |

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by maihuna » Sat Feb 12, 2011 9:29 pm
See B: 2n is div by as twice many no as n so if n=2, div by 1,2, 2*n = 4 is div by 1, 2, 4 not twice as many as n
if n=1, div by 1, 2*n=2, div by 1,2 so twice as many
if n=3, div by 1,3, 2*3 is div by, 1, 2, 3, 6 i.e. 4.

So B is suff.

Idea here is in 2*n, if n is even, then two scenario arises:
for n=2, 2*2 will have 3 divisor, 1,2,4
n=4, 2*4 will have 4 div, 1, 2, 4, 8

Idea is one factor 2 is common and so it never results in twice the divisor for 2*n.

but if n is odd, n will be having 2 divisor, namely 1 and n while 2*n will have, 1, 2, n, and 2*n. So B is suff
Charged up again to beat the beast :)