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Re: If a,b,k, and m are positive integers, is a^k a factor of [#permalink]

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12 Sep 2016, 11:48

Baten80 wrote:

If a,b,k, and m are positive integers, is a^k a factor of b^m?

1) a is a factor of b

2) k ≤ m

I don't know OA.

1) b^m = a^k * n. Since a is factor of B, let a =2, b=8 so 2^3m = 2^k * n. 2^k can only be a factor of 2^3m if the exponents of 2^k is less than 2^3m for all k and m. So not sufficient

2)k <= m, not sufficient because a and b relationship not given. If a = 3, b = 5, can be yes or no for a^k a factor of b^m

1+2) given a is a factor of b and the exponent of a will be smaller than or equal to the exponent of b, a^k is then a factor of b^m. C

If a,b,k, and m are positive integers, is a^k a factor of b^m?

1) a is a factor of b

2) k ≤ m

I don't know OA.

Question: is \(a^k\) a factor of \(b^m\) --> \(a^kx=b^m\), where \(x\) is an integer? --> \(x=\frac{b^m}{a^k}\). So basically the question is: Is \(x\) an integer \(>0\)?

(1) \(a\) is a factor of \(b\) --> \(ay=b\) --> \(x=\frac{a^my^m}{a^k}\) --> \(x=a^{m-k}y^m\). Now if \(m<k\) and \(a\) is not a factor of \(y\), then \(x\) will not be an integer. Not sufficient.

Or even without any algebra: if a and b are equal to say 3 and m<k (there are less b's than a's) then a^k won't be a factor of b^m. Though if k<=m then even if a and b are not equal still a^k will be a factor of b^m as there will be enough b's for a's.

(2) \(k\leq{m}\), not sufficient on it's own.

(1)+(2) \(x=a^{m-k}y^m\) and \(k<m\), hence \(x\) is an integer. Sufficient. (Or again as there are more b's then a's (enough b's for a) then a^k is a factor of b^m, for example (bbb)/(aa))