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

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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.

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

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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))

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

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Bunuel 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.

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

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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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If a, b, k, and m are positive integers, is a^k a factor of b^m?  [#permalink]

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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.

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

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Bunuel 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.

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
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Originally posted by Kinshook on 08 Sep 2019, 22:17.
Last edited by Kinshook on 09 Sep 2019, 03:27, edited 2 times in total.
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Re: If a, b, k, and m are positive integers, is a^k a factor of b^m?  [#permalink]

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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?

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.
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