Question

If $n$ is an odd natural number, then number of zeros at the end of ${99}^n+1$ is

• A. $2$
• B. $n$
• C. $2n$
• D. none of these

Answer 1

The correct option is A $2$

${99}^n+1$
$=(100-1{)}^n+1$
$={100}^n-{}^n{C}_1{100}^{n-1}+...-1{}^n{C}_n+1$ ...(since n is odd)
$={100}^n-{}^n{C}_1{100}^{n-1}+...-1^{n-1}{}^n{C}_{n-1}100$
$=100m$ where m is a positive integer.
Hence the number of zeros at the end will be 2.