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# How many digits are possible in the square of a number with ‘$n$’ digits.(This question has multiple correct options)A). $2n+1$B). $2n-1$C). ${{n}^{2}}$D). $2n$

Last updated date: 06th Sep 2024
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Hint: Before solving this question, we must know about perfect squares and square roots.
PERFECT SQUARE: If we multiply the number $x$ with itself then the result is called Perfect Square.
For example: $3~\ \ \times \ \ 3\text{ }=\text{ }9$ . Hence, $9$ is a perfect square.
SQUARE ROOT: If we get a number $x$ when we multiply a value $y$ with itself. Then we can say that the value $y$ is the square root of the value of $x$.
For Example, we have $9=3\times 3$, here we can say that $3$ is the square root of $9$

Complete step-by-step solution:
From the above results, we can choose the option $\left( d \right)$ as the correct option.
$\text{Example }1.2:-$
In our second example, consider the number ‘$25$’ which has ‘$2$’ digits, hence the value of ‘$n$’ is ‘$2$’
The square of $25$ is ‘$625$’, which is a three-digit number.
Now, substitute the value of ‘$n$’ in the given option to find the correct option.
\begin{align} & \left( a \right)\to \\ & \left( b \right)\to \\ & \left( c \right)\to \\ & \left( d \right)\to \end{align}\begin{align} & 2n+1=2\left( 2 \right)+1=5 \\ & 2n-1=2\left( 2 \right)-1=3 \\ & {{n}^{2}}={{2}^{2}}=4 \\ & 2n=2\left( 2 \right)=4 \end{align} Here we got the correct option as \left( b \right)2n-1 \text{Example }1.3:- In our third example, consider the number ‘625’ which has ‘3’ digits, hence the value of ‘n’ is ‘3’ The square of 625 is ‘$\text{390,625}$’, which is a Six-digit number. Now, substitute the value of ‘n’ in the given option to find the correct option. \begin{align} & \left( a \right)\to \\ & \left( b \right)\to \\ & \left( c \right)\to \\ & \left( d \right)\to \end{align}\begin{align} & 2n+1=2\left( 3 \right)+1=7 \\ & 2n-1=2\left( 3 \right)-1=5 \\ & {{n}^{2}}={{3}^{2}}=9 \\ & 2n=2\left( 3 \right)=6 \end{align}
Here we got the correct option as $\left( d \right)2n$

Note: From the above observations we can clearly say that for an even digit number the square of the number contains $2n-1$digits and for an odd digit number the square of that number contains $2n$ digits. While coming to the problem it is possible that the correct options are more than two.