Let $A=2012^{2012}$, B the sum of the digits of $A$, C the sum of the digits of $B$, and D the sum of the digits of $C$.

What is the value of $\mathrm{D}$?
On a $2012 \equiv 5(\bmod 9)$ hence $2012^{2012} \equiv 5^{2012}(\bmod 9)$. We therefore study the powers of 5 modulo $9: 5^{2} \equiv 7$ so $5^{3} \equiv-1$ and $5^{6} \equiv 1$. Now, we can write $2012=$ $2+6 \times 335$, hence:

$$
2012^{2012} \equiv 5^{2012} \equiv 5^{2} \times\left(5^{6}\right)^{335} \equiv 5^{2} \equiv 7 \quad(\bmod 9)
$$

thus $\mathrm{D} \equiv \mathrm{C} \equiv \mathrm{B} \equiv \mathrm{A} \equiv 7(\bmod 9)$.

On the other hand, $A<\left(10^{4}\right)^{2012}=10^{8048}$ so $A$ has at most 8048 digits, so $B \leqslant 9 \times 8048=72432$, so $\mathrm{B}$ has at most 5 digits, so $\mathrm{C} \leqslant 45$, and $\mathrm{D} \leqslant 13$. The only possibility is therefore $\mathrm{D}=7$.