UPSC CDS 1 2023 Math Paper

Concept:The last digit of a power depends on the cyclicity of the base's last digit.Explanation:For $9^{27}$, the last digit of $9^n$ is $9$ when $n$ is odd and $1$ when $n$ is even.Since $27$ is odd, the last digit of $9^{27}$ is $9$.For $27^{9}$, only the last digit of the base matters, which is $7$.The last digit of $7^n$ repeats every $4$: $7, 9, 3, 1$.Find the remainder when the exponent $9$ is divided by $4$: $9 \div 4$ gives remainder $1$.So the last digit of $27^{9}$ is the first in the cycle, which is $7$.Add the two last digits: $9 + 7 = 16$.The last digit of $16$ is $6$.Answer:$6$