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# k is a positive integer and 225 and 216 are both divisors of

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k is a positive integer and 225 and 216 are both divisors of  [#permalink]

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19 Feb 2012, 17:57
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Difficulty:

35% (medium)

Question Stats:

73% (02:00) correct 27% (02:12) wrong based on 317 sessions

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k is a positive integer and 225 and 216 are both divisors of k. If k=(2^a)*(3^b)*(5^c), where a, b and c are positive integers, what is the least possible value of a+ b+ c?

A. 4
B. 5
C. 6
D. 7
E. 8

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k is a positive integer and 225 and 216 are both divisors of  [#permalink]

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19 Feb 2012, 20:56
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4
vix wrote:
k is a positive integer and 225 and 216 are both divisors of k. If k=(2^a)*(3^b)*(5^c) , where a, b and c are positive integers, what is the least possible value of a+ b+ c?

A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Make prime factorization of 225 and 216: $$225=15^2=3^2*5^2$$ and $$216=6^3=2^3*3^3$$.

Now, in order to minimize the sum of the powers of k's primes, we should minimize k itself. Minimum value of k will be the least common multiple of 225 and 216 (since k is a multiple of both): $$LCM(225, 216)=k=2^3*3^3*5^2$$ --> $$a+b+c=3+3+2=8$$.

For more in this check Number Theory Chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.
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Re: k is a positive integer and 225 and 216 are both divisors of  [#permalink]

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01 Oct 2012, 03:07
2
Lets make factorization of 225 and 216 ..

225 = 5 X 5 X 3 X 3 X 3

216 = 2 X 2 X 2 X 3 X 3 X 3

K would have to have 3 two's , 225 has 3 threes and and so does 216 but they can be the same three threes so we count them only once ... 225 has 2 fives ... So we had them together and we get 3 + 3 + 2 = 8 (answer) ...
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Re: k is a positive integer and 225 and 216 are both divisors of  [#permalink]

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12 Nov 2017, 08:16
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: k is a positive integer and 225 and 216 are both divisors of &nbs [#permalink] 12 Nov 2017, 08:16
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