Sum of digits is so dangerous

If $C$ is the sum of the digits of $B$ , then $C \equiv B \equiv A \equiv 4444^{4444}$ modulo $9$ . Now $4444 \equiv 7 \pmod{9}$ and $7^3 \equiv 1 \pmod{9}$ and $4444 \equiv 1 \pmod{3}$ , so that $4444^{4444} \equiv 7^{4444} \equiv 7^1 \equiv 7 \pmod{9}$ Now $\log_{10} 4444^{4444} \approx 16210.70788$ , so that $4444^{4444} \approx 5.1 \times 10^{16210}$ is a $16211$ - digit number with leading digit $5$ , and hence $A \le 5 + 16210 \times 9 = 145895$ . But then $B \le 1 + 5\times9 = 46$ , and hence $C \le 4+1\times9 = 13$ . Since $1 \le C \le 13$ and $C \equiv 7 \pmod{9}$ , we deduce that $C = \boxed{7}$ .