Let A be the sum of the digits of 2012 ${ }^{2012}$. We define B as the sum of the digits of A, and $\mathrm{C}$ as the sum of the digits of $B$. Determine $C$.
We start by crudely bounding C. Since $2012^{2012} \leqslant 10000^{2012}$, it has fewer than $4 \times 2012 + 1 < 10000$ digits, so $A < 9 \times 10000 = 90000$. Similarly, $B < 9 \times 5 = 45$ and finally $C \leqslant 13$.

Now we look modulo 9 because we know that the operation of taking the sum of the digits leaves this congruence invariant. Thus,

$$
2012^{2012} \equiv 5^{2012} \equiv 5^{2} \equiv 7 \pmod{9}
$$

Thus, $C=7$.