2006 AMC 10B Problem 11

The tens digit depends only on the sum modulo $100$.

For $n \ge 10$, the product $n!$ includes the factors $5$, $10$ (and plenty of $2$s), so $100 \mid n!$. Every term from $10!$ onward contributes $0$ modulo $100$ and can be discarded.

That leaves $7! + 8! + 9!$. Rather than computing the large numbers, factor out $7! = 5040$:
$$7!\,(1 + 8 + 72) = 5040 \cdot 81.$$
Modulo $100$, $5040 \equiv 40$ and $81 \equiv 81$, so the sum is $\equiv 40 \cdot 81 = 3240 \equiv 40 \pmod{100}$.

The last two digits are $40$, so the tens digit is $4$.

The answer is $\boxed{\textbf{(C)}\ 4}$.

Questions about a specific digit are questions about a residue: tens digit means mod $100$. Factorials grow to absorb any fixed modulus quickly, so a long factorial sum always reduces to a handful of initial terms. Factoring the smallest factorial out of the survivors makes the leftover arithmetic tiny.

Direct addition: $5040 + 40320 + 362880 = 408240$, whose tens digit is $4$.

• Assuming factorials end in 00 as soon as they contain a single factor of 10 (from 5!), and so ignoring 7!, 8!, 9!.
• Reading the units digit ($0$) or the hundreds digit ($2$) instead of the tens digit.