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Expert replies
Source: — Data Sufficiency |

by stop@800 » Fri Oct 17, 2008 12:08 am
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by stubbornp » Fri Oct 17, 2008 1:23 am
oa B....curious for explanation...
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by stop@800 » Fri Oct 17, 2008 2:51 am
A
x=y^3 and n > y

= (x/y)^n
= y^2n

and n>y
so we can not get any relation for relating it to 1000


B
x > 5y

we need to check for
(x/y)^n = (x/y)^n
(x/y)^n < (5y/y)^n
(x/y)^n < 5^n

now as x and y are +ve integers
so min y can be 1
hence min x has to be 6

and if n > x
so n will certainly be greater than 6

and we all know 5 to the power a no greater than 6 will certainly be greater than 1000

hence B is sufficient

Hope this helps!!!
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