# Digit in units place of $1!+2!+\cdots+99!$

There isn't much I can add to the question description to expand upon the title. I came across this in a multiple choice test. The options were $3$, $0$, $1$ and $7$. I am absolutely stumped. Any pointers? By the way I ran a computer program and the answer is $3$. The rest of the test was quite easy, apparently I am missing something obvious.

• What is the units digit of $5!$? Feb 24 '14 at 8:56

Hint: If $n\geq 5$, what is the last digit of $n!$ ?

• Well this is embarrassing. Got it. :p
– Guy
Feb 24 '14 at 8:57

you need to add only $1!+2!+3!+4!$ which results in last digit as $3$, since number more than $5!$ or $5!$ itself, will have last digit as $0$, due to having a factor $10$.

Notice that: $$1!=1, \\ 2!=2, \\ 3!=6, \\ 4!=24, \\ 5!=120, \\ 6!=6 \times 5!=6 \times 120=720,$$ note that for any $n \geq 5$, $n!$ ended with zero, so it only suffices to sum up $1!+2!+3!+4!$ which is ended by $3$.

• Have you seen the already given answers? Apr 13 '18 at 16:45
• nop i solved question by myself, then answered for what i understood. Apr 13 '18 at 17:00
• @jane Welcome to MSE. Great that you started answering question. This answer you gave, however, is very similar to the ones already given. This is not very helpfull. Please don't be discouraged, but take some time to read a little about how this site opperates. For instance this, or the tour. Also, invest some time in learing to use MathJaX. We use it to format our maths on this site. Apr 13 '18 at 18:31