Question

Find the square root of $2116$ by prime factorization method.

Answer 1

Hint: In this question we have been given the number $2116$ for which we have to find the square root using the prime factorization method. We will solve this question by expressing the number as a product of prime numbers and whenever there are two same numbers present, we will consider only one of the factors and get the required solution.

Complete step by step answer:
We have the number given to us as $2116$
We can get the prime numbers of the number as:
$\begin{align}
  & 2\text{ }\left| \!{\underline {\,
  2116 \,}} \right. \\
 & 2\text{ }\left| \!{\underline {\,
  1058 \,}} \right. \\
 & 23\left| \!{\underline {\,
  529 \,}} \right. \\
 & 23\left| \!{\underline {\,
  23 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, we can write:
$\Rightarrow 2116=2\times 2\times 23\times 23$
Now we can see that the number $2$ is present two times and the number $23$ is present two times. This means that $2116$ is a perfect square an\d has a square root.
Now to find the square root, we will only consider one instance of the two, of both the factors present. therefore, we get:
$\Rightarrow \sqrt{2116}=2\times 23$
On multiplying the terms on the right-hand side, we get:
$\Rightarrow \sqrt{2116}=46$, which is the required square root of the given number.
Therefore, the square root of $2116$ is $46$, which is the required solution.

Note: In this question we had to find the square root of the given number. The cube root of a number can also be found using the prime factorization method, the only difference being that instead of considering one of two instances of the same factor, we need to consider one of three instances from the given factor. If there are factors which are remaining after selecting all the necessary instances, the number is not a perfect square and it doesn’t have an integer square root.