diegoasaavedra 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




answer is C, could somebody explain me this?

Thanks

(1) a is a factor of b
Let a=2, b=6
2 is a factor of 6
2^2 (= 4) is a factor of 6^2 (= 36)
2^3 (=8 ) is not a factor of 36
INSUFFICIENT

(2) k<=m
Does not tell us anything about a and b.
(You can also test this using numbers).
INSUFFICIENT.


Combining
a is a factor of b and k<=m
a^k = a*a*a*a... k times
b^m = b*b*b*b... m times

b^m/a^k = (b*b*b*b...m times)/(a*a*a*a... k times)
For every b in the numerator, there is atleast one a in the denominator (k<=m)
Since a is a factor of b, all those a's in the denominator cancel out. Hence, b^m/a^k is an integer.
SUFFICIENT.

Answer is (C).

If a^k is a factor of b^m then b^m is a multiple of a^k, thus b^m/a^k is an integer. So the question is asking if b^m/a^k is an integer.

statement 1 says a is a factor of b....if a is a factor of b then b is a multiple of a so calling b a function of a....
b = xa
where x is a positive integer

b^m/a^k = (xa)^m/a^k
= (x^m)(a^m)/a^k
= (x^m)a^(m-k)

in order for this term to be an integer a^(m-k) has to be an integer. The only way this is true is if m>=k.

Statement 2 says that k<=m (m>=k) so now we know for sure that b^m/a^k is an integer and thus, a^k is a factor of b^m....C