# Squares with Prime Factors

Find the smallest square that has at least $3$ different prime factors.

I tried finding the LCM of the $3$ smallest primes, $2$, $3$ and $5$, which is $30$, and $30^2$ equals $900$. But is this the right answer, and if yes, is there an easy way to work out this problem?

(I'm only a Year 7 (age $11-12$ for any non-UK people out there), so please explain clearly how you found the solution and if you have a formula for working out this problem, please make it so I can understand it.)

• Should the prime factors be all different? otherwise 16 is an answer Mar 22, 2017 at 8:31
• @user35508 yes they should I'll put that in the question
– bio
Mar 22, 2017 at 8:31
• Well, the only "gotcha" is that they didn't say 3 distinct prime factors. If the primes are required to be distinct, then your answer is correct. If they're not required to be distinct, then the answer is ... Mar 22, 2017 at 8:32
• @bio....but $8$ is not a perfect square Mar 22, 2017 at 8:36
• yes fair point I deleted the comment
– bio
Mar 22, 2017 at 8:37

Squaring a number doesn't change which primes are factors, and a smaller (positive) number will have a smaller square, so you need to find the smallest number which has three different prime factors, then square it. For any given three primes, the smallest number which has those primes as factors is their LCM, which (because they are prime, and different) is just their product. So choosing smaller primes will give you a smaller answer, and therefore the best thing to do is take the three smallest primes, to get $(2\times3\times5)^2$. If the question had said "four distinct prime factors", it would be $(2\times3\times5\times7)^2$, and so on. (Similarly, if it had asked for the smallest cube with three distinct prime factors, the answer would be $2\times3\times5)^3$.)