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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