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If k^3 is divisible by 240 what is the least possible value

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If k^3 is divisible by 240 what is the least possible value [#permalink]  26 Apr 2012, 02:14
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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
[Reveal] Spoiler: OA
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Re: Different ways to solve the problem : If K^3 is divisible [#permalink]  26 Apr 2012, 02:24
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sugu86 wrote:
If K^3 is divisible by 240, wat is the least possible value of integer k?

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

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

Make prime factorization: 240=2^4*3*5.

Now, all primes of k^3 must be multiples of 3 (because of 3), so the least value of k^3 so that it's divisible by 2^4*3*5 is k^3=2^6*3^3*5^3 --> k=2^2*3*5=60.

Hope it's clear.
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]  29 May 2012, 06:30
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hi Buneal your works are just grate

k^3=2^6*3^3*5^3
how power of 2 got 6

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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]  29 May 2012, 06:37
marzan2011 wrote:
hi Buneal your works are just grate

k^3=2^6*3^3*5^3
how power of 2 got 6

Posted from my mobile device

We have that:
(i) k^3 is divisible by 2^4*3*5;
(ii) primes of k^3 must be multiples of 3;

Now, k^3 is divisible 2^4, so the power of 2 in k^3 must be multiple of 3 and more that 4: the least multiple of 3 which is more that 4 is 6.

Hope it's clear.
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Re: If k^3 is divisible by 240 what is the least possible value   [#permalink] 29 May 2012, 06:37
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