What is the smallest counting (\[n\]) number that would make \[756n\] a perfect square?

Hint: When a number is multiplied by itself then it is called the square of the number. Here, we will first factorize the given number i.e., \[756\] by using the prime factorization method. Then we will find the number which is not paired. The number so obtained will be the smallest counting (\[n\]) number that would make \[756n\] a perfect square.

Complete step by step answer:
We have to find the smallest counting (\[n\]) number that would make \[756n\] a perfect square.
Factoring the given number i.e., \[756\] using prime factorization method, we get
\[ \Rightarrow 756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7\]
On pairing, we get
\[ \Rightarrow 756 = \left( {2 \times 2} \right) \times \left( {3 \times 3} \right) \times 3 \times 7\]
Now as we can see, \[3\] and \[7\] are not paired.
Since, in order to make the given number a perfect square we need to make the pair of each of the factors. Hence, we need to make a pair of \[3\] and \[7\].
So, we need to multiply \[\left( {3 \times 7} \right)\] i.e., \[21\] in the given number.
On multiplying \[756\] with \[21\], we get \[15876\].
\[ \Rightarrow \sqrt {15876} = 126\]
So, we can see that on multiplying \[21\] with \[756\], we are getting a perfect square.
Therefore, \[21\] is the smallest counting (\[n\]) number that would make \[756n\] a perfect square.

Note:
The square root of a perfect square is always a whole number. Also, prime factorization is the method of factoring a number in the multiples of prime numbers. Mathematically, \[a \times a = {a^2}\] is the square of any real number \['a'\] or in other words, a perfect square number \[{a^2}\] is the result of the product of \['a'\] and \['a'\].