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. 2
• B. 3
• C. 4
• D. none of these

Answer 1

Answer: (A)

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