If n = k^2pr^3 where k, p and r are different prime numbers what is

2 ways of doing this.

Method 1: If we want the lowest possible value of n, then the prime numbers have to be 2, 3 and 5.

Then we have 6 ways of putting the values of k, p and r and we get to the answer. A little application would help in the fact that the lowest value should be given to the number with the highest power. So if we take r = 2, k = 3 and p = 5, we get n = 9 * 5 * 8 = 720

Method 2:

Break each of the Options into its factors to see which of them fits our requirements.

Option A: 180 = \(2^2 * 3^2 *5\). This is not possible as we need a cubic power.


Option B: 240 = \(2^4 * 3 *5\). This is not possible.


Option C: 300 = \(2^2 * 3 *5^2\). This is not possible as we need a cubic power.



Option D: 360 = \(2^3 * 3^2 *5\). This Fits our requirement.


No need to check Option E as it is greater than 360.



Option D

Arun Kumar