I only want 3.

The number of such integers is the number of non-negative integer solutions of $a_1 + a_2 + a_3 + ... + a_{2013} = 3$ , which are ${{3+2013-1}\choose 3} = \boxed{1361529455}$ in number. The bijection is created from the idea that a general number of the form is $\overline{a_1a_2a_3...a_{2013}} [a_i \geq 0]$ , if any $a_i$ at the left is zero, the number has less digits.

There are three possibilities- the number is made of 0's and one 3, the number is made of 0's and one 1 and one 2, or the number is made of three 1's and 0's. There are 2013 places to put a digit in.

1. To define a number that is made out of 0's and one 3, we shall choose the decimal place (from 1 to 2013) in the number which contains the digit 3. There are "2013 choose 1"=2013 ways to do it.
2. To define a number that is made out of 0's and one 1 and one 2, we shall first choose two decimal places (from 1 to 2013) in the number which contain the digits 1 and two. There are "2013 choose 2"= $\frac{2013 \times 2012}{2}$ ways to do it, now we need to choose the place, out of the two we chose, which contains the digit 1. There are "2 choose 1"=2 ways to do it. For summary, there are $2013 \times 2012$ numbers that fill the conditions.
3. To define a number that is made out of 0's and 3 1's, we shall choose 3 decimal places (from 1 to 2013) in the number which contain the digit 1. There are "2013 choose 3"= $\frac{2013 \times 2012 \times 2011}{3}$ ways to do it.

The sum of those three possibilities, by the law of sum, is 1361529455. Thereforethe number's last three digits are 455.

Sorry, my English is poor.