Question

For integer $n>1$, the digit at units place in the number ${\sum }_{r=0}^{100}r!+2^{2^n}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

Since the digit at units place in each of $5!,6!,...,100!$ is $0!+1!+2!+3!+4!=34$.

Therefore the digit at unit place in $\sum _{r=0}^{100}r!$ is $4$.

Now, $2^{2^n}=2^{4k},k\in N$ $(2^n$ is a multiple of $4$ form $n>1)$

$\therefore$ The digits at units place in $2^{2^n}=2^{4k}={16}^k$ is $6$.

Thus, the digits at units place in $\sum _{r=0}^{100}r!+2^{2^n}$ is $0$.