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If k^3 is divisible by 240 what is the least possible value [#permalink]
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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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hi Buneal your works are just grate

would you please explain
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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marzan2011 wrote:
hi Buneal your works are just grate

would you please explain
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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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.
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Re: Different ways to solve the problem : If K^3 is divisible [#permalink]
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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 \(k^3=2^6*3^3*5^3\) --> \(k=2^2*3*5=60\).

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?
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]
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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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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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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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
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If k^3 is divisible by 240 what is the least possible value [#permalink]
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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
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Re: If k^3 is divisible by 240 what is the least possible value [#permalink]
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sugu86 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


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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If k^3 is divisible by 240 what is the least possible value [#permalink]
given,
\(k^3\) is divisible by 240 which means \(k^3\) is a multiple of 240
i.e. \(k^3 = 240 * (non zero integer 'n') \)
\(k^3= n* 24*10 =n*8*3*2*5 = n*2^4 *3*5 \)
the least integer value of n will give least integer value of k, to make the above expression a perfect cube the least integer value of n is
\(n=2^2 *3^2* 5^2, thus, k^3=2^6*3^3*5^3\)
therefore least possible value of k is:
\(2^2*3*5=60\)
ANSWER: C
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If k^3 is divisible by 240 what is the least possible value [#permalink]
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Given that k^3 is divisible by 240 and we need to find the minimum value of k

\(\frac{k^3}{240}\). Now we have \(k^3\) in the numerator so all terms in the numerator will have a power which is a multiple of 3

=> to find the least value of k we need to find the least value of \(k^3\) and to do that we need to factorize 240 and make all powers of the prime factors of 240 as a power of 3

240 = \(2^4 * 3^1 * 5^1\)

For all powers to be multiples of 3 we can make them as \(2^6 * 3^3 * 5^3\)

=> \(k^3 = 2^6 * 3^3 * 5^3\) = \((2^2 * 3 * 5)^3\)
=> k = \(2^2 * 3 * 5\) = 4 * 15 = 60

So, Answer will be C
Hope it helps!
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