unit digit of series containing factorial

Question

UNIT DIGIT

The unit digit of the following expression $$ (1!)^{99!} + (2!)^{98!} + (3!)^{97!} + (4!)^{96!} + \ldots + (99!)^{1!} $$

iam getting unit digit as 1 + 6 + 6 + 6 + 0 +-----+0 = 9 as answer am i correct

Answer 1

I felt a little bit confusing about the derivation at first sight but I can say it is perfectly right.

The simple calculations are as follows:

$\bullet$ Since $98!~ mod4 = 0$, $2^{98!} mod10 =6$, it is like the cycle of $2 \rightarrow 4 \rightarrow 8 \rightarrow 6 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 6$.

$\bullet$ The unit digit of every power of 3!, which is 6, is 6.

$\bullet$ Since $96!~ mod2 = 0$, $(4!)^{96!} mod10 =6$, it is like the cycle of $4 \rightarrow 6 \rightarrow 4\rightarrow 6$.

$\bullet$ From $5!=120$ on, every factorial has its unit digit 0, so any power of them should have unit digit 0, which confirms your result.