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.)
Yes, that's the right answer, and it's also the best way of working it out in my opinion.
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 $2\times3\times5)^3$.)