prady2231 wrote:

Find the sum of the digit of the least number K such that 2K is a square and 3K is a cube.

1. 9

2. 8

3. 13

4. 15

6. 16

Take it one step at a time.

2k is a square. A square needs to have even powers of every prime factor. So k must already have one or three or five or ... 2s.

The smallest acceptable value of k would be 2. k could also be \(2^3\) or \(2^5\) etc.

2*k = 2*2 is a perfect square. If k has another prime factor, it must ALREADY exist in even power.

3K is a cube. A cube needs to have all powers of prime factors as multiples of 3.

So k must have two or five or eight ... 3s.

The smallest acceptable value of k would be 3*3

3*k = 3*3*3 is a perfect cube. Any other prime factor it may have must ALREADY be in power of 3. So the power of 2 in k must be 3.

So we see that k must have at least three 2s and two 3s.

\(k = 2^3 * 3^2 = 72\)

Sum of the digits = 7+2 = 9

_________________

Karishma

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