Find the last digit of the sum $1!+2!+3!+...+49!$ Is there any formula for finding the last digit of the factorials?
How to approach these type of questions?
Thanks in advance.
 A: First of all, when you see this sort of seemingly intractable problem, don't despair. There's usually a very simple "trick" that makes the problem trivial.
In this case, you have to realise two things:
1) only the sum of last digits contributes to the last digit of the final sum.
2) factorials of larger numbers have a lot of zeroes at the end.
So your problem reduces to deciding the final term you have to consider. Luckily this is a very easy problem. Because:
$5! = 120$
$6! = 720$
and so forth, every factorial after that ending with a zero.
So you only have to consider the sum $1! + 2! + 3! + 4!$.
Even that's simplified by recognising that $3!$ ends with a $6$ and $4!$ with a $4$, so they will sum up to give $0$ as the last digit.
Turns out all you have to consider is $1! + 2!$, which is just $3$.
I wanted to put an exclamation point at the end of the last line to emphasise how easy the whole thing was, but decided not to because it might look like a factorial! :)
A: well as the comments stated, you should observe that for $n \geq 5$ we have $n!$ is a multiple of $5$. For example $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 24 \times 5$$
and $$6 ! = 6 \times 5! = 6 \times 24 \times 5$$
and so on. Now you should know that any multiple of $5$ is either $0$ or $5$ for instance $5,10,15,20,25,.........$ as you can see.
now $$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$$
and $5! + 6! + ..........+49!$ is a mutliple of $5$ as we said.
However, $$5! + 6! + ..........+49!$$ is an even number because they are also a multiple of $2$ 
And so we have $1! + 2! + 3! + 4!$ is odd 
and $5! + 6! + ..........+49!$ is even
and you should know that  odd + even = odd. $$(2k + 1) +2m = 2k +2m + 1 = 2(k +m) + 1$$ which is odd.
And so the sum is odd.
and so $5! + 6! + ..........+49!$ ends with $0$ because it is a multiple of $5$ and an even number.
and $33$ + a number that ends with $0$ is will result in a number with last digit = $3$
