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

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

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 greater than 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 ≤ 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))


Answer: C.
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good question
#1
a is a factor of b
a = 3 and b = 6
so
if k>m ; we get no and when k<m we get yes
test with k=1 and m= 1 and k=3 & m=1
insufficient
#2
k<=m
again relation of a & b is not clear so it can be yes and no
a=3 and b = 6 or a =2 and b = 3 so values of m & k wont make any difference insufficeint

from 1 & 2
yes sufficient to say that a=3,k=1& b=6 ,m=2 sufficient to say that a^k a factor of b^m
IMO C

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?
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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)
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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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