2006 AMC 10B Problem 11

1. The tens digit of a sum only depends on the last two digits of each number. So look for terms that end in $00$ and ignore them.
2. $10! = 1 \cdot 2 \cdot 3 \cdots 9 \cdot 10$ contains the factor $10$ and also the factors $2$ and $5$, which make another $10$. So $10!$ is a multiple of $100$ and ends in $00$.
3. Every later factorial, like $11! = 11 \cdot 10!$, is a multiple of $10!$, so it also ends in $00$. All the terms from $10!$ to $2006!$ add nothing to the last two digits.
4. That leaves only three terms. Work them out:
$$7! = 5040,\qquad 8! = 8 \cdot 5040 = 40320,\qquad 9! = 9 \cdot 40320 = 362880.$$
5. Only the last two digits matter: $40 + 20 + 80 = 140$, which ends in $40$. So the whole sum ends in $40$.

The tens digit is $4$.

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

A question about the last digit or the tens digit only cares about the last one or two digits of each number. Factorials quickly pick up lots of factors of $2$ and $5$, so from some point on they all end in zeros and drop out. That turns a sum with two thousand terms into a sum of just a few.

Add the three numbers in full: $5040 + 40320 + 362880 = 408240$. Its tens digit is $4$.

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 $\boxed{\textbf{(C)}\ 4}$.