Find the smallest natural number 'n' which becomes a perfect square when divided by 3 and a perfect cube when divided by 5.

Sure, I can help with that.

First, let's understand the problem. We are looking for a natural number 'n' which when divided by 3 becomes a perfect square and when divided by 5 becomes a perfect cube.

Let's denote the perfect square as 'a^2' and the perfect cube as 'b^3'. So, we have:

n/3 = a^2 n/5 = b^3

From these equations, we can express 'n' as:

n = 3a^2 = 5b^3

Since 'n' must be the same in both equations, we can set them equal to each other:

3a^2 = 5b^3

To find the smallest 'n', we need to find the smallest values of 'a' and 'b' that satisfy this equation.

The smallest perfect square (other than 0) is 1 (which is 1^2), and the smallest perfect cube (other than 0) is 1 (which is 1^3). However, if we plug 'a' = 1 and 'b' = 1 into the equation, we get 3 ≠ 5, which is not true.

The next smallest perfect square is 4 (which is 2^2), and the next smallest perfect cube is 8 (which is 2^3). If we plug 'a' = 2 and 'b' = 2 into the equation, we get 12 ≠ 40, which is also not true.

If we continue this process, we find that 'a' = 5 and 'b' = 3 satisfy the equation (since 75 = 75). Therefore, the smallest 'n' is 3*5^2 = 75.