Question

The remainder left out when $8^{2n}-(62{)}^{2n+1}$is divided by 9 is :

• A. 2
• B. 7
• C. 8
• D. \N

Answer 1

The correct option is A 2

$(8{)}^{2n}-(62{)}^{2n+1}=(64{)}^n-(62{)}^{2n+1}=(63+1{)}^n-(63-1{)}^{2n+1}=\left[{}^n{C}_0\right(63{)}^n+{}^n{C}_1(63{)}^{n-1}+{}^n{C}_2(63{)}^{n-2}+...{+}^n{C}_{n-1}\left(63\right){+}^n{C}_n]-\left[{}^{2n+1}{C}_0\right(63{)}^{2n+1}{-}^{2n+1}{C}_1\left(63{)}^{2n}{+}^{2n+1}{C}_2\right(63{)}^{2n-1}-....+(-1{)}^{2n+12n+1}{C}_{2n+1}]=63\times \left[{}^n{C}_0\right(63{)}^{n-1}+{{}^{n}C}_1(63{)}^{n-2}+{{}^{n}C}_2(63{)}^{n-3}+....]+1-63\times \left[{}^{2n+1}{C}_0\right(63{)}^{2n}{-}^{2n+1}{C}_1(63{)}^{2n-1}+....]+1$
$\Rightarrow 63\times$ some intergral value +2
$\Rightarrow 8^{2n}-(62{)}^{2n+1}$ when divided by 9 leaves 2 as the remainder.