Problem - 4261
Let $N=4568^{7777}$, $a$ be the sum of digits in $N$, $b$ be the sum of digits in $a$, and $c$ be the sum of digits in $b$. Find $c$.
Because $N=4568^{7777} < 10000^{7777}= 10^{31109}$, therefore $N$ has at most $31109$ digits. It follows that $a < 9\times 31109=27998$. Then, we find $b < 2 + 5\times 9 = 47$, and $c < 4 + 9 = 13$.
Meanwhile, we have $$c\equiv 4568^{7777}\equiv 5^{7777} \equiv (5^3)^{2592}\times 5\equiv (-1)^{2592}\times 5\equiv 5\pmod{9}$$
Therefore, we find $c=\boxed{5}$.