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

Hi Bunuel GMATPrepNow VeritasKarishma EMPOWERgmatRichC

Is my below understanding correct w.r.t. Bunuel 's solution above:
1. An integer raised to a positive integer MUST to be an integer.
2. An integer raised to a negative integer may / may not be an integer.
e.g. 2^ (-1) is not an integer, whereas 1^ (-1) is an integer.

This is where St 2 really plays an important role. Am I correct?
Can I not play with picking nos in such problems with given restrictions in qs stem
since I can not choose negative no and 0?

Hi adkikani,

TESTing VALUES works on lots of different Quant questions (in both PS and DS) and it can certainly be used on this prompt. After doing a little work, you might actually discover a Number Property in this prompt (although you don't need to use that rule to get the correct answer).

We're told that A, B, K and M are all POSITIVE INTEGERS. We're asked if A^K is a factor of B^M. This is a YES/NO question.

1) A is a factor of B.

IF....
A=B=2 and K=M=1.... 2^1 = 2 is a factor if 2^1 = 2 and the answer to the question is YES
A=B=2 and K=2, M=1.... 2^2 = 4 is NOT a factor if 2^1 =2 and the answer to the question is NO
Fact 1 is INSUFFICIENT

2) K ≤ M.

IF....
A=B=2 and K=M=1.... 2^1 = 2 is a factor if 2^1 = 2 and the answer to the question is YES
A=3, B=2 and K=M=1.... 3^1 = 3 is NOT a factor if 2^1 = 2 and the answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we know...
A is a factor of B.
K ≤ M

Since A is a factor of B, we know that A divides evenly into B. In that situation, the only way for A^K to NOT be a factor of B^M would be when K is greater than M (you may have already noticed that in the work that we did above). Since K must be less than/equal to M, A^K will always be a factor of B^M and the answer to the question is ALWAYS YES.
Combined, SUFFICIENT.

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Given: a, b, k, and m are positive integers

Asked: Is a^k a factor of b^m?

(1) a is a factor of b.
Even if a is a factor of b, If k>m, then a^k may or may not be a factor of b^m
NOT SUFFICIENT

(2) k ≤ m.
Since relation between a and b is unknown
NOT SUFFICIENT

Combining (1) & (2)
(1) a is a factor of b.
(2) k ≤ m.
a^k will be a factor of b^m since b = a n
\(b^m = a^m * n^m = a^k a^{m-k} * n^m\)
a^k is factor of b^m
SUFFICIENT

IMO C

Yes, an integer raised to a positive integer will be an integer. Since a positive integer exponent is just the base multiplied by itself as many times as specified in the power,

\(4^5 = 4*4*4*4*4\) - it will remain an integer.

An integer raised to a negative power may or may not be an integer. Correct as shown by you in your examples.

I did the question by thinking of number plugging only here. It seems easy that way.

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

Is \(\frac{b^m}{a^k} = \frac{b*b*b*b* ...}{a*a* ...}\) an integer?

This will be an integer when all a's get cancelled by the b's above i.e. if it were something like this \(\frac{15*15*15*15}{5*5*5}\).

(1) a is a factor of b.

This is necessary but not sufficient. 5 is a factor of 15 in example above. But what if we have four 15s but six 5s? Then all 5s will not cancel out.

(2) k ≤ m.

This helps but is not sufficient. What if k is 3 as shown in the example above but instead of 5s in denominator, we have 100s? Then we will not get an integer.

Using both, a is a factor of b and we have fewer a's. So it will be something similar to the fraction above and we will get an integer.
Answer (C)

Given: a, b, k, and m are positive integers

Target question: Is a^k a factor of b^m?

Statement 1: a is a factor of b.
This information seems partially useful. However, since we know nothing about the exponents, a^k and b^m can be practically any magnitude.
To see what I mean consider these two possible cases that satisfy statement one:
Case a: a = 2, b = 4, k = 1 and m = 2. In this case, a^k = 2^1 = 2, and b^m = 4^2 = 16. In this case, the answer to the target question is YES, a^k IS a factor of b^m
Case b:a = 2, b = 4, k = 5 and m = 1. In this case, a^k = 2^5 = 32, and b^m = 4^1 = 4. In this case, the answer to the target question is NO, a^k is NOT a factor of b^m
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT


Statement 2: k ≤ m
Since we have no information about a and b, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: if a is a factor of b, then we can also say that b is a multiple of a.
So, let's say b = ax (for some positive integer)
Statement 2: k ≤ m
So, let's say m = k + n, where n is some integer greater than or equal to 0 (this satisfies the condition that k ≤ m).

This means, we can write: b^m = (ax)^(k + n) = [(ax)^k][(ax)^n]
= [(a^k)(x^k)][(a^n)(x^n)]
= (a^k)[(x^k)(a^n)(x^n)]
Since we can express b^m as a multiple of a^k, we can be certain that a^k IS a factor of b^m
The answer to the target question is YES, a^k IS a factor of b^m

Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

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