A Brilliant "Sum of Digits" Problem

Let $DS(n)$ be the digit sum of $n$ . Certainly $4444^{4444} \equiv 7 \pmod{9}$ . Now $\lfloor4444\log_{10}4444\rfloor = 16210$ , and so $DS(4444^{4444}) \le 16211 \times 9 = 145899$ . Thus $DS(DS(4444^{4444})) \le 1 + 5 \times 9 = 46$ , and hence $1 \le DS(DS(DS(4444^{4444}))) \le 4 + 9 = 13$ . The only number between $1$ and $13$ which is congruent to $7$ modulo $9$ is $\boxed{7}$ .

$4444^{4444}=(493\times 9+7)^{(3\times 1481+1)}$ . So

$4444^{4444}\equiv 7^1\equiv 7 \pmod 9$ .

Since the remainder when a number is divided by $9$ is the same as that when the sum of digits of that number is divided by $9$ , the answer is $\boxed 7$ .