100

This is a sum of an arithmetic progression .

$\begin{aligned} S & = 1+2+3+...+100 \\ & = \sum_{k=1}^n k & \small \color{#3D99F6}{n=100} \\ & = \frac {n(n+1)}2 \\ & = \frac {100(100+1)}2 \\ & = 5050 \end{aligned}$

The last digit of the sum is $\boxed{0}$ .

The answer the is 5,050 so the last digit is 0

very easy , by using formula n(n+1)/2 (100*101)/2 =10100/2 =5050 and the last digit is 0

n(n+1)/2 = 100(100+1)/2

             = 100 *101/2

             = 50 *101

             = 5050

Therefore the end digit is 0.

i used the formula of sum of n terms of A.P. \frac{n}{2}*{2a+(n-1)d}
where n= total number of terms(here n=100) a=1st term of the series(here a=1) d=common difference between 2 consecutive terms(here d=1)

Putting all the values in the formula we get- \frac{100}{2} {2 1+(100-1)1} =5050