Too Many Terms For A Last Digit

1! = 1 (units digit 1) 2! = 2 (ud 2) 3! = 6 (ud 6) 4! = 24 (ud 4) 5!, 6!.... the units digit will be 0 because all the terms will include 5x2=10

units digit of 1!+2!+... will be 1+2+6+4+0+0+0.... = 3

1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

Then, 5! = 120

From 5 onwards all positive integers have factorials ending with 0.

So, the last term will definitely be 3.

$n!$ is defined as: $n!=n\times (n-1)\times(n-2)\times\cdots 3\times 2\times 1$ Now observe that if $n\geq5$ : $n!=n\times (n-1)\times(n-2)\times\cdots \color{#3D99F6}{5}\times 4 \times 3\times \color{#D61F06}{2}\times 1$ Those 5 and 2 multiply to give 10.And any number multipied by 10 will have last digit 0.Hence: $\boxed{n!\; \text{will have last digit 0 iff}\; n\geq 5}$ Now we can clearly see that all the factorials from $5!$ upto $100!$ will have last digit 0,which implies that their sum also has last digit 0.Hence we just need to compute the last digit of $1!+2!+3!+4!=33$ hence the last digit is $\boxed{3}$

5! Is 120 the last digit of 120 is 0 4! Is 24 the last digit of that is 4 so we add that to the last digit of 3!=6 2!=2 and 1!=1 so we do 4+6+2+1 witch is 13 and the last digit of 13 is 3 so the answer is 3

Look at residues mod10.

(1!) mod10 = 1
(2!) mod10= 2
(3!) mod10=6
(4!) mod10=4
(5!) mod10=0

since 6!=6*5! and 5! mod10=0, we have that 6! mod10=0. The same holds for every term until 100!. Thus all we have to sum are:

1+2+6+4 = 3mod10.

So the units digit is 3.

For k > 4 , unit digit of k! = 0 so the 1!+2!+3!+4! = 33 and unit digit is 3.

1!+2!+3!+4! = 33

After that every factorial contain the product of 2 and 5 therefore is going to finish by 0 so the last digit is 3.