Question

The last digit of the number ${32}^{32}$ is

• A. $4$
• B. $6$
• C. $8$
• D. None of these

Answer 1

The correct option is B

$6$

The explanation for the correct answer:

Solve for the last digit of ${32}^{32}$

${32}^{32}={(2+3\times 10)}^{32}$

$=2^{32}+{10}^k$, where $k$ belongs to$N$.

So, the last digit in ${32}^{32}=$the end digit in $2^{32}$

The cyclicity of $2$is $4$ That means the unit’s digits of $2^x\left[x>0\right]$ follow the pattern of digits $2,4,8,$ and $6$

The power in the problem is $32$ When we divide the power by cyclicity, we get a remainder
$\Rightarrow \frac{32}{4}$

Remainder $=0.$

The remainder $0$ represents that we have to take the last number in the cyclicity pattern which becomes the required answer $=6$

Hence, option (B) is the answer.