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

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

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New post 28 Dec 2009, 11:04
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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 is less than or equal to m.

Correct answer: both statements together are sufficiant but neither statemement alone is sufficient

1) a is a factor of b
a = Lb => a^k = (Lb)^k = L^k * b^k
and ???

Thanks for your help!

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

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New post 28 Dec 2009, 11:49
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nailgmattoefl wrote:
Hello,

Here is the problem:

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 is less than or equal to m.

Correct answer: both statements together are sufficiant but neither statemement alone is sufficient

1) a is a factor of b
a = Lb => a^k = (Lb)^k = L^k * b^k
and ???

Thanks for your help!

See you!


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 \(>0\)?

(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) \(k\leq{m}\), 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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Re: Math Question Algebra  [#permalink]

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New post 03 Mar 2010, 05:37
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nailgmattoefl wrote:
Hello,

Here is the problem:

" If a,b,k and m are positive integer, is a^k a factor of b^m

1) a is a factor of b

2) k < m

Correct answer: both statements together are sufficiant but neither statemement alone is sufficient

1) a is a factor of b
a = Lb => a^k = (Lb)^k = L^k * b^k
and ???

Thanks for your help!

See you!


given: a,b,k,m > 0 and are Integers

Ques: Is \(\frac{b^m}{a^k}\) = Integer

S1: \(\frac{b}{a}\) = Integer. Not Suff... as m and k could be anything.
S2: m > k .. Not Suff as no info on b and a.

S1 + S2: SUFF as b is divisible by a and m>k.. Hence the result would always be an integer
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Re: Math Question Algebra  [#permalink]

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New post 28 Dec 2009, 12:04
Thanks very Bunuel !!!

I got it!

3 months will not be too much to reach 700+ level.

That's crazy how you forgot mathematics when you are no more using it !!!

See ya!
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Re: Math Question Algebra  [#permalink]

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New post 03 Mar 2010, 08:55
amoore wrote:
Can someone make a more thorough explanation? I'm having trouble understanding (1).

thanks


For stmt 1 to be enough to answer whether a^k is factor of b^m,
k should be less than m or else there can be case when k > m then k-m>0 so there the final stmt will be 1/a^(k-m) which is not a integer hence 1 not enough to answer.
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Re: Math Question Algebra  [#permalink]

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New post 04 Mar 2010, 01:00
Select smart numbers, plug them, and you can easliy conclude that statement 1 alone is not sufficient.

Only on combining with statement 2, do we get the answer.
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Re: If a, b, k, and m are positive integers, is a^k a factor of  [#permalink]

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New post 16 Jan 2012, 01:36
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 is less than or equal to m

SOLUTION:
(1) a is a factor of b.
example: 4 is a factor of 16 & 20.
so, 4^2 is a factor of 16^1 but not 20^1.

Hence, Not sufficient

(2) k is less than or equal to m
we don't know anything about a,b. so we cant say.

NOT Sufficient

On combining also,
example : a=4;b=16,20;k=2;m=1
4^2 is a factor of 16^1 but not 20^1

Hence, NOT sufficient.

ANS E

hope it will help. :)
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Re: If a, b, k, and m are positive integers, is a^k a factor of  [#permalink]

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