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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 \(k^3=2^6*3^3*5^3\) --> \(k=2^2*3*5=60\).

would you please explain k^3=2^6*3^3*5^3 how power of 2 got 6

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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.

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

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08 May 2014, 06:22

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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 \(k^3=2^6*3^3*5^3\) --> \(k=2^2*3*5=60\).

Answer: C.

Hope it's clear.

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

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 \(k^3=2^6*3^3*5^3\) --> \(k=2^2*3*5=60\).

Answer: C.

Hope it's clear.

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

Sorry don't follow you...

You mean that the least value of k must be 2^2*3^2*5^2? Why?
_________________

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

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04 Jan 2016, 17:21

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2

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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.

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.
_________________

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

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06 Jan 2016, 07:17

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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 \(k^3=2^6*3^3*5^3\) --> \(k=2^2*3*5=60\).

Answer: C.

Hope it's clear.

Dear Bunuel How to get k=2^2*3*5=60[/m] From \(2^4*3*5\) is \(k^3=2^6*3^3*5^3\) --> \(k=2^2*3*5=60\). Appreciate your reply Many thanks

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

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18 Feb 2016, 05:52

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.

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

Since k^3/240 = integer, we can say that the product of 240 and some integer n is equal to a perfect cube. In other words, 240n = k^3.

We must remember that all perfect cubes break down to unique prime factors, each of which has an exponent that is a multiple of 3. So let’s break down 240 into primes to help determine what extra prime factors we need to make 240n a perfect cube.

240 = 24 x 10 = 8 x 3 x 2 x 5 = 2 x 2 x 2 x 3 x 2 x 5 = 2^4 x 3^1 x 5^1

In order to make 240n a perfect cube, we need two more 2s, two more 3s and two more 5s. Thus, the smallest perfect cube that is a multiple of 240 is 2^6 x 3^3 x 5^3.

To determine the least possible value of k, we can take the cube root of 2^6 x 3^3 x 5^3 and we have:

2^2 x 3 x 5 = 4 x 3 x 5 = 60

Thus, the minimum value of k is 60.

Answer: C
_________________

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