integer problem, plz help!

We can write the question mathematically like so:

Does there exist an integer x, such that:
x(a^k) = b^m?
(You get this from the definition of a factor. For $ to be a factor of #, there is some integer such that $*integer = #.)

Now let's take a look at what statement 1 tells us. "a is a factor of b." Thus, there is some integer y such that ay = b. Plugging that into the equation given by the question, we have:

xa^k = b^m
xa^k = (ay)^m
xa^k = (a^m)*(y^m).
Now, y^m is just some integer, so we can cut down on confusion by calling that z.
xa^k = za^m
Dividing both sides by a^k gives us:
x = za^(k-m)

Now, the question is still, is x an integer? x will be an integer if k>=m, because then a^(k-m) will be an integer, and z is an integer. However, if k<m, then a^(k-m) will not be an integer because an integer to a negative power is less than 1. In that case, x will not be an integer. So, because we don't know from statement 1 if k>=m, we do not have enough information and statement 1 is not sufficient.

Considering statement 2, we have:
xa^k = b^m, and k<=m. We can express x as:
x = (b^m)/(a^k).

We could have a,b,m, and k such that:
x=(3^2)/(2^1) = 9/2 (not an integer)

Or, we could have a,b,m, and k such that:
x=(4^2)/(2^1)=16/2=8 (an integer)

So, statement 2 doesn't give us enough information to determine if x is an integer.

Combining the statements, we find that we had reduced statement 1 to "x will be an integer if k>=m." Statement 2 gives us this very piece of information. So, put together we have sufficient information and the answer is C.

Hi ,

Am i missing something..
In this statement
xa^k = za^m
Dividing both sides by a^k gives us:
x = za^(k-m)

when we divide will it not be x=za^(m-k).

Please let me know.. thanks in advance .