multiple

[ketkoag]

Is b^m the multiple of a^k?
1) b is a multiple of a
2) m>k

IMO : E please confirm the answer..

[cramya]

I would say C

is b^m = some integer * a^k

Stmt I

b = k*a

Is (ka) ^m = some integer * a^k

Is k^m * a^m / a^k = some integer

We dont know since m>k or m<k (k^m is always an integer)

INSUFF

Stmt II

m>k

No idea about a and b

INSUFF

Together:

Is k^m * a^m / a^k = some integer

yes
a^m / a^k is an integer since m>k so k^m (int) * a^m / a^k (int) = integer

[rahulg83]

I simply plugged in few values..
Obviously none of them alone is sufficient
Suppose a=5, b=15 (as per st I)
St II m>k, lets take m=3, k=2

only when m=1, k=2 will b^m will not be a multiple of a^k, u can check will all the other values

But what if m=-1 and k=-2 (we don't know whether m and k are positive)

b^m = 1/15; a^k = 1/25, 1/15 is not a multiple of 1/25

hence E

Please somebody correct me if i am wrong anywhere..