Question

The last digit of $3^{3^{4n}}+1,n\in N$, is

• A. ${}^4{C}_3$
• B. ${}^8{C}_7$
• C. $8$
• D. $4$

Answer 1

Answer: (A), (D)

The correct options are
A ${}^4{C}_3$
D $4$

$3^{4n}$ $=(3^4{)}^n$
$={81}^n$
$=(1+80{)}^n$
$=1+[{}^n{C}_{1}80+{}^n{C}_2{80}^2+{}^n{C}_3{80}^3...+{}^n{C}_n{80}^n]$
$=1+\lambda$
Taking $80$ common from $\lambda$ we get
$=1+80k$ where $k$ is a constant...(i)
$3^{1+80k}$ $=3\left(3^{80k}\right)$
$=3\left(9^{40k}\right)$
$=3\left({81}^{20k}\right)$
$=3\left(\right[1+80{]}^{20k})$
$=3(1+\alpha 80)$ ...similar to (i)
$=3+240\alpha$
Hence $3^{3^{4n}}+1$
$=3+240\alpha +1$
$=240\alpha +4$
$=10\left(24\alpha \right)+4$
$=10\beta +4$
Hence the last digit is $4$.