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 ?

Say $x_i, i=1,\ldots,15$ are 15 3-digit numbers. The probability that 10 divides their sum is equal to the probability that
$$\sum_{i=1}^{15} x_i=0 \pmod{10}$$
But this corresponds to choosing 15 1-digit numbers (zero's and repeats allowed).
For each $x_i$ the probability that $x_i=z_j \pmod{10}$ (for some $z_j\in\{0,\ldots,9\}$) is $\frac{1}{10}$. Then you have to compute the probability that $10$ divides $\sum_{i=1}^{15}x_i\pmod{10}=\sum_{i=1}^{15}z_i$.
The sum of 15 1-digit numbers ranges from $0$ to $9\times 15=135$. Within this range there are 14 numbers which give you zero modulo 10. Those numbers are $\{0,10,\ldots,130\}$. If you find how many ways are there to choose 15 1-digit numbers such that their sum equals one of these numbers, you're done.
