<s>[INST] What is the smallest positive integer n such that $2^5 · 3 · 5^2 · 7^3 ·$ $n$ is a perfect square Is there some kind of rule that I need to solve this? Can someone give me some clue how to solve this? Thanks
My teacher gave the solution as $42$.
Can someone explain why?
 [/INST] I will use the rule that:
$$a^2b^2=(ab)^2$$
We want all exponents to be even. Why? Let's use an example.
(Assume all variables are integers, and are not perfect squares themselves).
$a^2b^4$ is a perfect square because it equals $(ab^2)^2$. But $a^2b^3$ is not a perfect square because there is no way to convert it into something of the form $(xy)^2$. The best we can do is $(ab)^2\times b$, but $b$ is not in the squared term, therefore it is not a perfect square (remember, we assumed $b$ is not a perfect square).
We want to find the value of $n$ in $2^5\times 3\times 5^2\times 7^3\times n$ so that the latter expression will be a perfect square.
First step: Attempt to make as many exponents even as possible. We can rewrite our expression as:
$$2^4\times 5^2\times 7^2\times 2\times 3\times 7\times n$$
We can now group the first three terms into a perfect square.
$$(2^2\times 5\times 7)^2\times 2\times 3\times 7\times n$$
$$=140^2\times 2\times 3\times 7\times n$$
We need to find the lowest value of $n$ such that $2\times 3\times 7\times n$ is a perfect square. It is easily seen that the value of $n$ is $2\times 3\times 7$, which is $42$. Therefore:
$$\color{green}{\boxed{n=42}}$$
Hope I helped!

P.S. If you want to find out what happens when $n=2\times 3\times 7$, read on.

When $n=2\times 3\times 7$:
$$140^2\times 2\times 3\times 7\times n=140^2\times 2^2\times 3^2\times 7^2$$
$$=140^2\times (2\times 3\times 7)^2$$
$$=140^2\times 42^2$$
$$=(140\times 42)^2$$
$$=5880^2$$
$$=34574400$$
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