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Find the smallest positive integer $k$, such that product of $420$ and $k$ is a perfect square.

Please help me in this question.

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What you really need is not an answer to the specific question but an understanding of the general principle you need to be deploying here. So let's think about the general problem: Given $n$ find the smallest $k$ such that $m = n \cdot k$ is a perfect square. What to do?

Good hint for questions such as this: think about factorization (as the question, is $m$ a perfect square is a question about how it factorizes). So ...

What are the properties of the factors of perfect squares $m$? Well since $m = j^2$ any factor of $j$ must occur twice in $m$. So any prime factor of $m$ must occur an even number of times.

And that gives us our clue to how to proceed:

Step one. Factorize $n$.

Step two. Look to see which primes don't appear an even number of times in the factorization.

Step three. Take $k$ to be the product of those primes with an odd number of occurrences in $n$.

So -- by brute force construction -- $n \cdot k$ is the product of primes, each one of which now occurs an even number of times as we want. So $m = n \cdot k$ is a perfect square. And we've put into $k$ the fewest prime factors needed to make the product with $n$ a perfect square. So we are done.

Now you have the principle of the thing, you should be able to see how to do this problem too: Find the smallest number $k$ such that $420 \cdot k$ is a perfect cube and other problems in the same family.

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Observe that the power of $3,5$ and $7$ in $420$ is $1$

So, to make the power of $3$ even, we need to multiply $420$ by $3^{2a+1}$ where $a$ is a non-negative integer.

So, the general value of $k$ is $3^{2a+1}\cdot5^{2b+1}\cdot7^{2c+1}$ where $a,b,c$ are non-negative integers.

So, the smallest positive integer value of $k$ is $3\cdot5\cdot7=105$ (putting $a=b=c=0$)

Observe that $\frac{420}{105}=4=2^2$ is also a perfect square.

If $A\cdot B$ is perfect square, so will be $\frac AB$ if $B\mid A$

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