If $n$ is a positive integer, the symbol $n$ ! (read " $n$ factorial") represents the product of the integers from 1 to $n$. For example, $4 !=(1)(2)(3)(4)$ or $4 !=24$. The ones (units) digit of the sum $1 !+2 !+3 !+4 !+5 !+6 !+7 !+8 !+9 !+10$ ! is
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
We note that

$$
\begin{gathered}
1 !=1 \quad 2 !=(1)(2)=2 \quad 3 !=(1)(2)(3)=6 \\
4 !=(1)(2)(3)(4)=24 \quad 5 !=(1)(2)(3)(4)(5)=120
\end{gathered}
$$

Thus, $1 !+2 !+3 !+4 !+5 !=1+2+6+24+120=153$.

Now for each positive integer $n \geq 5$, the ones digit of $n$ ! is 0 :

One way to see this is to note that we obtain each successive factorial by multiplying the previous factorial by an integer. (For example, $6 !=6(5 !)$.)

Thus, if one factorial ends in a 0 , then all subsequent factorials will also end in a 0. Since the ones digit of 5 ! is 0 , then the ones digit of each $n$ ! with $n>5$ will also be 0 . Alternatively, we note that for each positive integer $n$, the factorial $n$ ! is the product of the positive integers from 1 to $n$. When $n \geq 5$, the product represented by $n$ ! includes factors of both 2 and 5 , and so has a factor of 10 , thus has a ones digit of 0 .

Therefore, the ones digit of each of 6!, 7!, 8!, 9!, and 10! is 0 , and so the ones digit of $6 !+7 !+8 !+9 !+10$ ! is 0 .

Since the ones digit of $1 !+2 !+3 !+4 !+5$ ! is 3 and the ones digit of $6 !+7 !+8 !+9 !+10$ ! is 0 , then the ones digit of $1 !+2 !+3 !+4 !+5 !+6 !+7 !+8 !+9 !+10$ ! is $3+0$ or 3 .

(We can verify, using a calculator, that $1 !+2 !+3 !+4 !+5 !+6 !+7 !+8 !+9 !+10 !=4037913$.)