Question

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

• A. $3$
• B. $4$
• C. $2$
• D. none of these

Answer 1

The correct option is C $2$

$(99{)}^n+1$
$=(100-1{)}^n+1$
$=1-(1-100{)}^n$ ...(since n is odd)
$=1-[1-{}^n{C}_{1}100+...(-1{)}^r{}^{100}{C}_r{100}^r+...]$
$=1-[1-100+100(k\left)\right]$ where K is a integer.
$=100-100k$
Hence the ending digits have 2 zeros, and k becomes a negative integer since ${99}^n+1$ is positive.