How can we find the least perfect square number divisible by each one of 3, 4, 5, 6,8 with easy 1 min solution?

Question

If we find the LCM of these numbers we get 120 but thats not the answer and after that I'm not able to do anything further, can anybody help me?

Answer 1

For the corrected version of the problem:

The number must be divisible by $120=2^3\times3\times5$. It is a perfect square. So it is at least $2^4\times3^2\times 5^2$.

Answer 2

For $p\mid n^2$ and $p$ prime it follows that $p\mid n$. Since $2,3,5$ are the primes in $3,4,5,6$, we need at least $2,3,5$ as divisors for $n$. So $n=30$ and $n^2=900$ is the minimal choice. If you add $8$, then $p=2$ has to appear with multiplicity $3$ in $n^2$, so at least with multiplicty $2$ in $n$. Hence $n=2\cdot 30$ is minimal, which means $n^2=3600$.

Answer 3

We can actually disregard $6$ and $8$, because we're already demanding that $n^2$ be divisible by $3$ and by $4$. So once we find our $n$, we'll see that it's also divisible by $9$ and by $16$.

So the least common multiple of $3$, $4$ and $5$ is $60$. This means the answer is $60^2 = 3600$.

Now, $900$ won't do because it is divisible by $4$ but not by $8$.