Question

If $S={99}^n+1,n>1,n\in N$, then which of the following is/are correct?

• A. When $n$ is odd, then the last two digits is $00$
• B. When $n$ is odd, then the last two digits is $02$
• C. When $n$ is even, then the last two digits is $02$
• D. When $n$ is even, then the last two digits is $22$

Answer 1

Answer: (A), (C)

The correct options are
A When $n$ is odd, then the last two digits is $00$
C When $n$ is even, then the last two digits is $02$

$S={99}^n+1=1+(100-1{)}^n=1+\{{}^n{C}_0{100}^n-{}^n{C}_1{100}^{n-1}+\cdots +(-1{)}^n\}$

When $n$ is odd, we get
$S=100k,k\in Z$
The last two digit is $00$

When $n$ is even, we get
$S=2+100k_1,k_1\in Z$
The last two digit is $02$