Question

Let n be an odd natural number greater than 1. Then the number of zeros at the end of the sum${99}^n+1$ is

• A. 3
• B. 4
• C. 2
• D. None of these

Answer 1

The correct option is C 2

$1+{99}^n=1+(100-1{)}^n=1+$
${\{}^n{C}_0{100}^n{-}^n{C}_1{.100}^{n-1}+.....{.}^n{C}_n\}$
Because n is odd $=100\{{}^n{C}_0{.100}^{n-1}-{{}^n{C}_1.100}^{n-2}+......{{}^{n}C}_{n-2}.100+{{}^{n}C}_{n-1}\}$
= 100 $\times$ integer whose units place is different from 0
[${\because }^n{C}_{n-1}=n$, has odd digit at unit place]
$\therefore$ There are two zeros at the end of the sum ${99}^n+1$