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
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Karishma
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