Question

What could be the possible One’s digits of the square root of each of the following numbers?

$\left(i\right)9801$

$\left(ii\right)99856$

$\left(iii\right)998001$

$\left(iv\right)657666025$

Answer 1

Lets observe the following squares,

$1^2=1$, digit in one’s place is $1.$

$2^2=4$, digit in one’s place is $4.$

$3^2=9$, digit in one’s place is $9.$

$4^2=16$, digit in one’s place is $6.$

$5^2=25$, digit in one’s place is $5.$

$6^2=36$, digit in one’s place is $6.$

$7^2=49$, digit in one’s place is $9.$

$8^2=64$, digit in one’s place is $4.$

$9^2=81$, digit in one’s place is $1.$

${10}^2=100$, digit in one’s place is $0.$

Now, lets evaluate the the possible One’s digits of the square root of each of the given numbers.

$\left(i\right)9801$

If the number ends with $1$, then the one’s digit of the square root of that number may be $1$ or $9$. Therefore, one’s digit of the square root of $9801$ is either $1$ or $9$.

$\left(ii\right)99856$

If the number ends with $6$, then the one’s digit of the square root of that number may be $4$ or $6$. Therefore, one’s digit of the square root of $99856$ is either $4$ or $6$.

$\left(iii\right)998001$

If the number ends with $1$, then the one’s digit of the square root of that number may be $1$ or $9$. Therefore, one’s digit of the square root of $998001$ is either $1$ or $9$.

$\left(iv\right)657666025$

If the number ends with $5$, then the one’s digit of the square root of that number will be $5$. Therefore, the one’s digit of the square root of $657666025$ is $5$.