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

Question

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?

Answer 1

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 $$(2\cdot 3\cdot 5)^6=30^6=729 000 000$$

Answer 2

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 $30^6$.