Problem 289

We note that $$1! = 1 \qquad 2! = (1)(2) = 2 \qquad 3! = (1)(2)(3) = 6$$$$4! = (1)(2)(3)(4) = 24 \qquad 5! = (1)(2)(3)(4)(5) = 120$$ 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 $\text{n gt;5}$ 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 $1!+2!+3!+4!+5!+6!+7!+8!+9!+10! = 4\,037\,913$.)