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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]

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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
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Re: Different ways to solve the problem : If K^3 is divisible [#permalink]

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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, the powers of each prime of $$k^3$$ must be a 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]

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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

Posted from my mobile device

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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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29 May 2012, 06:37
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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) the powers of each prime of $$k^3$$ must be a multiple 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: Different ways to solve the problem : If K^3 is divisible [#permalink]

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08 May 2014, 07:22
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Bunuel wrote:
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, the powers of each prime of $$k^3$$ must be a 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.

Hi Bunuel,
To divide K^3. don't we need . 2^2*3^2*5^2.
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Re: Different ways to solve the problem : If K^3 is divisible [#permalink]

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08 May 2014, 08:25
seabhi wrote:
Bunuel wrote:
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, the powers of each prime of $$k^3$$ must be a 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.

Hi Bunuel,
To divide K^3. don't we need . 2^2*3^2*5^2.

You mean that the least value of k must be 2^2*3^2*5^2? Why?
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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08 May 2014, 08:32
Hi Bunuel,

Because to convert $$240=2^4*3*5$$ to be divisible by $$k^3$$ don't we need $$2^2*3^2*5^2$$
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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08 May 2014, 08:35
seabhi wrote:
Hi Bunuel,

Because to convert $$240=2^4*3*5$$ to be divisible by $$k^3$$ don't we need $$2^2*3^2*5^2$$

We are told that k^3 is divisible by 240, not vise-versa. Also, what is 2^2*3^2*5^2? Is it the value of k, k^3, ...
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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08 May 2014, 13:18
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I just used 60 as an answer. Since this is a 600 level question, I guess using answer directly would make sense. Thoughts?

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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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26 Jun 2015, 10:11
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If k^3 is divisible by 240 what is the least possible value [#permalink]

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04 Jan 2016, 18:21
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zoskybuck wrote:
I really need help with the question in this attached photo..I already solved that the crude way because am good with integers raised to the powers of up to 5 or 6 but i know that there is a way to do that with prime factorization or something..I am looking for that way..The explanation on the Manhattan GMAT strategy guide is making my brain hurt lol...Any clear and splendid explanation for this problem will be greatly appreciated...I started reading for the GMAT two months ago,so am still brushing up on lots of stuffs..Thank you guys...

Follow posting guidelines (link in my signatures) especially searching for a question before posting a new thread. Additionally, type in the complete question with all the options and do not use pictures to stand for the questions.

Refer above for the solution. For this question, the most straightforward way is to use the options to your advantage.

You know that k^3 must be divisible by 240 ---> You get an INTEGER value when you calculate $$k^3/240$$ ---> this eliminates A) directly as 12 is not divisible by 5 (=a factor of 240).

For B ---> if k =30=2*3*5 ---> $$k^3 = 2^3*3^3*5^3$$ and $$240 = 2^4*3*5$$, you can clearly see that 1 '2' from 240 will not be cancelled out when you write $$k^3/240$$ giving you a NON INTEGER VALUE. Eliminate this option.

For C ---> if $$k =60=2^2*3*5$$ ---> $$k^3 = 2^6*3^3*5^3$$ and $$240 = 2^4*3*5$$, you can clearly see that all the factors of 240 will now be cancelled out when you write $$k^3/240$$ giving you an INTEGER value. This is the correct answer.

As the question was asking about the LEAST value, no need to check D and E.

Hope this helps.
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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04 Jan 2016, 21:53
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zoskybuck wrote:
I really need help with the question in this attached photo..I already solved that the crude way because am good with integers raised to the powers of up to 5 or 6 but i know that there is a way to do that with prime factorization or something..I am looking for that way..The explanation on the Manhattan GMAT strategy guide is making my brain hurt lol...Any clear and splendid explanation for this problem will be greatly appreciated...I started reading for the GMAT two months ago,so am still brushing up on lots of stuffs..Thank you guys...

Hi,
k^3 is div by 240..
since k is to the power of 3 we will have to prime factorize 240 in terms of 3rd power....
240 = 2^3*2*3*5....
this means k has to be div by each of above, base can be taken if anything to power of 3 is there as the other powers will get adjusted by the power of k, k^3..
since k is an integer, we cannot take k to be div by 3rd root of 3 and 5, it has to be div by 3 and 5..
so k^3 is div by 2^3*2*3*5..
remove the 3rd powers..
so k will be div by 2*2*3*5 or 60..

hope it helped you with concept.
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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06 Jan 2016, 08:17
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Bunuel wrote:
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, the powers of each prime of $$k^3$$ must be a 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.

Dear Bunuel
How to get k=2^2*3*5=60[/m] From $$2^4*3*5$$ is $$k^3=2^6*3^3*5^3$$ --> $$k=2^2*3*5=60$$.
Many thanks

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If k^3 is divisible by 240 what is the least possible value [#permalink]

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18 Feb 2016, 06:52
Dear Apple08

I assume you're doing the same mistake as seabhi. The question is asking for the "least possible value of the integer k". It's not asking for $$k^3$$.

Of course, to end up with a perfect cube ($$k^3$$) you need to add $$2^2 * 3^2 * 5^2$$ but thats not questioned. Rephrasing the question then you're actually seeking for the 3rd root of ($$k^3$$).
$$\sqrt[3]{2^6 * 3^3 * 5^3} = 2^2 * 3 * 5$$ and since $$2^2 * 3 * 5$$ equals 60, the correct answer is C.

Hope it helped you
Greetings from Germany
Dennis

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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]

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05 Aug 2017, 07:02
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Re: If k^3 is divisible by 240 what is the least possible value   [#permalink] 05 Aug 2017, 07:02
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