What is the smallest positive integer n such that $2^5 · 3 · 5^2 · 7^3 ·$ $n$ is a perfect square

Question

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?

Answer 1

Hint:

The simplest approach would be to consider how many times do you need to had 2,3,5, or 7 appear in $n$ so that the exponents are all even. Note that 5 already has an even exponent so it isn't doesn't have to be considered.

Consider that any perfect square will have even exponents in its prime factorization:

$4 = 2^2$

$9 = 3^2$

$36 = 2^2*3^2$

and so forth.

Answer 2

Remember that if the factorization is $\prod_i p_i^{n_i}$ then n^2 = $\prod_i p_i^{2n_i}$ so all the prime factors have to have even powers. In this case 2, 3, and 7 are uneven so the answer would be their product.

Answer 3

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