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Joined: 18 Dec 2009
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If a, b, k, and m are positive integers, is a^k a factor of [#permalink]
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28 Dec 2009, 12:04
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73% (00:40) correct 27% (00:38) wrong based on 438 sessions
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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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Joined: 02 Sep 2009
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If a, b, k, and m are positive integers, is a^k a factor of [#permalink]
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28 Dec 2009, 12:49
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 \(b\) > \(ay=b\) > \(x=\frac{a^my^m}{a^k}\) > \(x=a^{mk}y^m\). 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) \(x=a^{mk}y^m\) 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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28 Dec 2009, 13:04
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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Joined: 22 Dec 2009
Posts: 324

Re: Math Question Algebra [#permalink]
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03 Mar 2010, 06:37
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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03 Mar 2010, 09:55
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 km>0 so there the final stmt will be 1/a^(km) which is not a integer hence 1 not enough to answer.



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Re: Math Question Algebra [#permalink]
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04 Mar 2010, 02:00
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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16 Jan 2012, 02:36
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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