if a, b, k and m are positive integers is a^k factor of b...

galinaphillips wrote:if a, b, k and m are positive integers is a^k factor of b^m?

(1) a is a factor of b
(2) k less than or equal to m

both together are sufficient but neither alone is sufficient

I saw this at mba.com but am not sure why this is true?
Thanks!

Take b=6, a=2

1 - insufficient.
Take b = 6, a = 2. we know a is a factor of b. But a^2 is not a factor of b.
Take b = 8, a = 2, we know a is a factor of b. And a^2, a^3 are factors of b.

2 - insufficient. Without knowing relationship between a and b, we cannot
say for sure

1 and 2 together - sufficient.

If k<=m and a is a factor of b, then b^m / a^k will always be an
integer and hence a^k will be a factor of b^m.

Thanks a lot!

Hi , I am sorry but I didn't get it

if a = 2 , k = 6
b = 8 , m = 2

In this case a raised to power k is a factor of b raised to power m and k is more than m , how does 'k less than or equal to m' helps here.

Please advise.

Croyant wrote:Hi , I am sorry but I didn't get it

if a = 2 , k = 6
b = 8 , m = 2

In this case a raised to power k is a factor of b raised to power m and k is more than m , how does 'k less than or equal to m' helps here.

Please advise.

What if k were 10. Then it isn't a factor, thus insufficient

Question is

b^m=n*a^K?

For this to be true a must be factor of b ie b=ja...where j is integer
now b^m=n* j^k * b^k

now for that eq to be true k can be less than or equal to m.

So 'C'