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Need simpler logic!

by vinay1983 » Wed Sep 18, 2013 2:39 am
If k^3 is divisible by 240 what is the least possible value of integer K?

A. 12
B. 30
C. 60
D. 90
E. 120

The earlier answers were not understandable
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by vipulgoyal » Wed Sep 18, 2013 2:45 am
k^3/2*2*2*2*3*5
k must have two 2 one 3 and one 5, hence smallest value = 60
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by theCodeToGMAT » Wed Sep 18, 2013 2:57 am
240= 2 x 3 x 2 x 2 x 5 x 2 == 2 ^ 4 x 3 x 5

hence, we would need a number that contains .. atleast TWO 2, one 3 and one 5.

The reason for TWO 2's is that if there were only one Two than cubing "k" would create only 2^3; but we need at least 2^4

So, Answer = 60 [C]

Hope this solution is understandable; feel free to ask if you have any doubts!!

Njoy!

vinay1983 wrote:If k^3 is divisible by 240 what is the least possible value of integer K?

A. 12
B. 30
C. 60
D. 90
E. 120

The earlier answers were not understandable
Last edited by theCodeToGMAT on Wed Sep 18, 2013 2:59 am, edited 1 time in total.
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by sanju09 » Wed Sep 18, 2013 2:58 am
vinay1983 wrote:If k^3 is divisible by 240 what is the least possible value of integer K?

A. 12
B. 30
C. 60
D. 90
E. 120

The earlier answers were not understandable
240 = 2^4 times 3^1 times 5^1, and the smallest cube number divisible by 240 must have 2 more 2s, 2 more 3s, and 2 more 5s to become 2^6 times 3^3 times 5^3, whose cube root (k here) is [spoiler]2^2 times 3^1 times 5^1 = 60.

Pick C[/spoiler]
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by Brent@GMATPrepNow » Wed Sep 18, 2013 7:00 am
vinay1983 wrote:If k^3 is divisible by 240 what is the least possible value of integer K?

A. 12
B. 30
C. 60
D. 90
E. 120
The approach shown by vipulgoyal and rahulmittal87 is perfect, but if you didn't spot that approach, don't forget that we can also test the answer choices.

k^3 is divisible by 240
Since 240 = (2)(2)(2)(2)(3)(5), we know that k^3 MUST HAVE at least four 2's, one 3 and one 5 in its prime factorization.

Now check the answer choices.

A) k = 12 = (2)(2)(3)
Since there's no 5 in the prime factorization of k, k^3 cannot have at least one 5 in its prime factorization
ELIMINATE A.

B) k = 15 = (2)(3)(5)
So, k^3 = (2)(3)(5)(2)(3)(5)(2)(3)(5)
As we can see, k^3 does not have at least four 2's in its prime factorization
ELIMINATE B.

C) k = 30 = (2)(2)(3)(5)
So, k^3 = (2)(2)(3)(5)(2)(2)(3)(5)(2)(2)(3)(5)
Perfect! k^3 has at least four 2's, one 3 and one 5 in its prime factorization

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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