# Smallest positive integer that is divisble by 2, 3 and 5 and that is also a perfect square and perfect cube [closed]

My son is having this problem:

Smallest positive integer that is divisible by $$2$$, $$3$$ and $$5$$, and that is also a perfect square and perfect cube.

He was busy using Excel, but I thought: Isn't there a faster way?

• $$(2\cdot3\cdot5)^{6m}$$ Commented Nov 20, 2018 at 12:13
• @labbhattacharjee: Since he said "smallest", my answer would be $(2\cdot 3\cdot 5)^6$, Commented Nov 20, 2018 at 12:16

Since number has to be divisible by $$2,3,5$$ therefore it should be divisible by $$30$$. Since you want a perfect square and a perfect cube therefore your number should be of the form $$30^{6t}$$ . Where t is the integral number satisfying all such conditions. In your case $$t=1$$ , therefore answer is $$30^6$$.
If $$n$$ is an integer, then
• $$n^2$$ is a perfect square
• $$n^3$$ is a perfect cube
An integer that has both these properties is $$n^{2\cdot 3}=n^{6}$$, since $$n^{6}=\underset{\color{red}{\text{square}}}{\underbrace{(n^{3})^\color{red}{2}}}=\underset{\color{green}{\text{cube}}}{\underbrace{(n^{2})^\color{green}{3}}}$$ You also want this to be divisible by $$2,3,5$$, these are prime numbers so the smallest non-zero such number is $$(2\cdot 3\cdot 5)^6=30^6=729 000 000$$