16.1.5 ** The sum of the digits of $4444^{444}$ is $A$, and $B$ is the sum of the digits of $A$. Find the sum of the digits of $B$ (all numbers are in decimal).
Since the number of digits of $4444^{4444}$ does not exceed $4 \times 4444=17776$, we have $A \leqslant 17776 \times 9=$ 159984, so $B \leqslant 1+5 \times 9=46$, and thus the digit sum of $B$, $c \leqslant 4+9=13$.
Since a number is congruent to its digit sum $\bmod 9$, we have
$$
c \equiv B \equiv A \equiv 4444^{4444} \equiv 7^{4444}=\left(7^{3}\right)^{1481} \cdot 7 \equiv 1^{1481} \cdot 7 \equiv 7(\bmod 9),
$$

thus $c=7$. That is, the digit sum of $B$ is 7.