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If \(k^3\) is divisible by 240 what is the least possible

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by GMATGuruNY » Sat Nov 23, 2019 3:18 am
VJesus12 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

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT
We can PLUG IN THE ANSWERS, which represent the smallest possible value of k.
Since the question stem asks for the smallest possible value, start with the smallest option.
When the correct answer is plugged in, (k*k*k)/240 = integer.

A --> (12*12*12)/240 = (12*12)/20 = noninteger
Eliminate A.

B --> (30*30*30)/240 = (30*30)/8 = (15*30)/4 = noninteger
Eliminate B.

C --> (60*60*60)/240 = (60*60)/4 = 15*60 = integer

The correct answer is C.
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by Scott@TargetTestPrep » Thu Dec 12, 2019 6:38 pm
VJesus12 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

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT
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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