Sum of random numbers is divisible by $10$ Suppose that $15$ three-digit numbers have been randomly chosen and we are about to add them. What is the probability that the sum would be divisible by $10$?
If there were only two or three random numbers we could enumerate the cases in which last digit comes out to be $0$ and hence calculate probability but for $15$ numbers that seems messy so is there a smart way to do it 
Edit:
I have tried another approach which finds the possible sums of $15$ three-digit numbers and then find the sums divisible by $10$ in the same range. So I get:
Number of Sums divisible by $10$ in $[1500,14985]=1349$
Total Number of Sums in $[1500,14985]=13486$
And then $P=\frac{1349}{13486}$, but as a comment suggests that this approach does not cater for the fact that a sum may be reached in a multiple of ways. So how can we cater for this fact? 
I am guessing may be multinomial can be of help ?

Add up the first 14 numbers. Then, whatever the remainder of the result modulo $10$, adding the 15th number will give each of the possible remainders with equal probability. So the answer is $0.1$.
