Would Factoring 180 Help?

Relevant wiki: Perfect Squares, Cubes, and Powers

$180=2^2×3^2×5$ .

Since the product is a perfect cube, each of the exponents must be a multiple of 3.

Therefore the least value of $N$ is $2×3×5^2=\boxed{150}$ .

Given, The product of 180 and a positive integer N is a perfect cube. Let $180\times N \space = \space x^3$ $=> \space 2^2\times3^2\times5^1\times N=x^3$ There for the product to be a perfect cube, The value of $N$ must be $2^1\times3^1\times5^2$ = $150$

• First convert the original number to its factors. 180=2 2 3 3 5
• Then you have to complete series of three numbers. So because there are two 2s, use one more 2; because there are two 3s, use one more 3; and because there is one 5, use two more 5s.
• Then multiply the numbers that you used 2 3 5*5=150
• At the end you'll have the product of 180 150, and because it's made only by series of three numbers (2 2 2, 3 3 3 and 5 5*5) you now that is a perfect cube.