Question

For any integer $n>1,$ which of the following is/are correct?

• A. The unit place digit of $\sum _{r=0}^{100}r!$ is $4$
• B. The unit place digit of $2^{2^n}$ is $6$
• C. The unit place digit of $\sum _{r=0}^{100}r!+2^{2^n}$ is $0$
• D. The unit place digit of $\sum _{r=0}^{100}r!+2^{2^n}$ is $8$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The unit place digit of $\sum _{r=0}^{100}r!$ is $4$
B The unit place digit of $2^{2^n}$ is $6$
C The unit place digit of $\sum _{r=0}^{100}r!+2^{2^n}$ is $0$

The digit at unit place in each of $5!,6!,\dots ,100!$ is $0$ and
$0!+1!+2!+3!+4!=34$

Unit place digit in $\sum _{r=0}^{100}r!$ is $4$.
Now,
$2^n=4k(\because n>1)\Rightarrow 2^{2^n}=2^{4k}\Rightarrow 2^{2^n}=(16{)}^k$

So, the unit place digit in ${16}^k$ is $6$.

$\therefore$ The digit at unit place in ${\sum }_{r=0}^{100}r!+2^{2^n}$ is $4+6=10$
Hence, the unit place digit is $0$.