Question

A perfect square number can never have the digit …….. at the unit's place.
$\begin{align}
& A.1 \\
& B.4 \\
& C.8 \\
& D.9 \\
\end{align}$

Answer 1

Hint:
For this question, first write the definition of perfect square and its properties.
Observe all the properties and try to find out which given numbers are at units place.

Complete step by step solution:
To solve this question, first we have to know what is a perfect square?
Perfect square- A perfect square is a number that can be made by squaring a whole number.
Here are some properties are perfect square:
If a number has $1\,or\,9$ in the units place then its square ends in $1$.
If a number has $2\,or\,8$ in the units place then its square ends in $4$.
If a number has $3\text{ or }7$ in the units place then its square ends in $9$.
When a number ends with zero, its square ends with double zeros.
If a number ends with an odd number of zeros then it is not a perfect square.
In perfect squares, the digits at the units place are always $0, 1, 4, 5, 9$. The numbers having $2, 3, 8, 7$ are its end place are not perfect squares.
By the property (ii) it is clear that if we have $2, 3, 8, 7$ at the unit place then the number is not a perfect square.
So that we have only one option C that has a digit $8$ and if $8$ is at unit place then the given number is not a perfect number.

Hence option C is the correct one.

Note:
In such types of problems, students should know the properties of perfect squares otherwise they will face problems. Prime number is the whole number which has only two factors.