UPSC CDS 1 2023 Math Paper

Concept:The square of any odd integer is of the form $8k+1$, so any even power of an odd integer gives remainder 1 when divided by 8.Explanation:Let the odd integer be $2n+1$ (where $n$ is an integer and $2n+1>1$).$(2n+1)^2 = 4n^2+4n+1 = 4n(n+1)+1$.The product $n(n+1)$ is always even because two consecutive integers include one even number.Thus $n(n+1)=2m$ for some integer $m$, and $4 \times 2m = 8m$ is exactly divisible by 8.Therefore $(2n+1)^2 = 8m+1$, so it leaves remainder 1 when divided by 8.An even power of the odd integer is of the form $[(2n+1)^2]^k$ for some positive integer $k$.Since $(2n+1)^2 \equiv 1 \pmod{8}$, raising it to any exponent $k$ gives $1^k \equiv 1 \pmod{8}$.Hence every even power of every odd integer (greater than 1) leaves remainder 1 upon division by 8.Answer:1