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If k³ is divisible by 240

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If k³ is divisible by 240

by melguy » Mon Oct 31, 2016 7:05 pm
Hello

Please help me with the question below

If k³ 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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by crackverbal » Mon Oct 31, 2016 10:17 pm
Hi Melguy,

This is a good example for prime factorization question type in GMAT.

Here first we need to do is prime factorize 240.

240 = 2^4 * 3 * 5

Given k^3 is divisible by 240, so it has to be perfect cube.

To be a perfect cube, after prime factorizing a number the power of the primes should be divisible by 3,

So we have to multiply minimum 2^2 * 3^2 * 5^2 with 240, so that we get the power of primes divisible by 3.

So k^ 3 = 2^6 * 3^3 * 5^3 ,

So k = 2^2 * 3 * 5

K = 60.

We can multiply any multiples of 2, 3 and 5 where the power is divisible by three. But since question is asking least possible value of k, we should have multiply to 240 minimum 2 two's , 2 three's and 2 five's.

Another way of solving this question, use the answer choices.

Since question asks about least possible value of k, start with least in the answer choices.

A.K = 12. Here 12 ^ 3 is certainly not divisible by 240. Since it has no zero.
B.K = 15. Here 15 ^ 3 is certainly not divisible by 240. Again its not have a zero.
C.K = 60. Here 60 ^ 3 will divisible by 240. We can quickly prime factorize and check this. So the
answer is C.

Hope it is clear
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by fiza gupta » Tue Nov 01, 2016 6:51 am
k³ is divisible by 240
k is an integer
240 = > 2*2*2*2*3*5

k should be a perfect cube root of k³
k = 2*2*3*5 = 60
SO C
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by Scott@TargetTestPrep » Wed Nov 02, 2016 7:31 am
melguy wrote:
If k³ 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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by Matt@VeritasPrep » Fri Nov 11, 2016 4:24 pm
k³ / 240 = integer

k³ = 240 * integer

So 240 * integer must be a cube.

240 = 2 * 2 * 2 * 3 * 2 * 5

For that to be a cube, we need a multiple of 3 of each prime factor, since each prime factor must fit into each of the three roots.

We've got four 2s, so we need TWO more to get to a multiple of 3 of them.

We've got one 3, so we need TWO more again.

We've got one 5, so we need TWO more again.

That leaves us missing 2 * 2 * 3 * 5, so "some integer" = 2 * 2 * 3 * 5 = 60.

Going back to the beginning, we've got

k³ = 240 * 60

or

k³ = (2 * 2 * 3 * 5)³

or k = 60.
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