Let
		\[
		S = \sum_{i = 1}^{2012} i!.
		\]
		The tens and units digits of $S$ (in decimal notation) are $a$ and $b$, respectively. Compute $10a + b$.

[i]Proposed by Lewis Chen[/i]
1. We need to find the tens and units digits of the sum \( S = \sum_{i=1}^{2012} i! \).
2. Notice that for \( i \geq 10 \), \( i! \) ends in at least two zeros because \( 10! = 10 \times 9 \times 8 \times \cdots \times 1 \) includes the factors 2 and 5, which produce a trailing zero. Therefore, the factorials of numbers greater than or equal to 10 will not affect the last two digits of the sum \( S \).
3. Thus, we only need to consider the sum of the factorials from \( 1! \) to \( 9! \):
   \[
   S = 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9!
   \]
4. Calculate each factorial and sum them:
   \[
   \begin{align*}
   1! &= 1 \\
   2! &= 2 \\
   3! &= 6 \\
   4! &= 24 \\
   5! &= 120 \\
   6! &= 720 \\
   7! &= 5040 \\
   8! &= 40320 \\
   9! &= 362880 \\
   \end{align*}
   \]
5. Now, sum the last two digits of each factorial:
   \[
   \begin{align*}
   1! &\equiv 01 \pmod{100} \\
   2! &\equiv 02 \pmod{100} \\
   3! &\equiv 06 \pmod{100} \\
   4! &\equiv 24 \pmod{100} \\
   5! &\equiv 20 \pmod{100} \\
   6! &\equiv 20 \pmod{100} \\
   7! &\equiv 40 \pmod{100} \\
   8! &\equiv 20 \pmod{100} \\
   9! &\equiv 80 \pmod{100} \\
   \end{align*}
   \]
6. Sum these results modulo 100:
   \[
   01 + 02 + 06 + 24 + 20 + 20 + 40 + 20 + 80 = 213
   \]
7. The last two digits of 213 are 13. Therefore, the tens digit \( a \) is 1 and the units digit \( b \) is 3.
8. Compute \( 10a + b \):
   \[
   10 \cdot 1 + 3 = 13
   \]

The final answer is \(\boxed{13}\)