Question

Which of the following numbers is a perfect square?
${\text{A}}{\text{.}}$141
${\text{B}}{\text{.}}$196
${\text{C}}{\text{.}}$124
${\text{D}}{\text{.}}$222

Answer 1

Hint: In this question use the concept of prime factorization method, also we know that perfect squares have prime factors i.e. divisors of a number exist in pairs.

Complete step-by-step answer:

First number is $141$
Prime factorization: $141 = 3 \times 47$
$141$ has two divisors $3$ and $47$,both elements do not exist in pairs

So, $141$ is not a perfect square

Second number $196$
Prime factorization: $196 = 2 \times 2 \times 7 \times 7$
Its prime factors exist in pairs i.e. $\left( {2 \times 2} \right) \times \left( {7 \times 7} \right)$

So, $196$ is a perfect square

Third number is $124$
Prime factorization: $124 = 2 \times 2 \times 31$
First element $2$ exists in pair but $31$ is left out

So, $124$ is not a perfect square

Fourth number $222$
Prime factorization: $222 = 2 \times 3 \times 37$
All elements do not exist in pairs

So, $222$ is not a perfect square

Hence the answer is $196$

Note: In this type of question first we find out the prime factors of the given number then we arrange the factors in pairs, if all the elements exist in pairs then that number is a perfect square. We found that only $196$ has its prime factors in pairs so therefore it is a perfect square.