Find numbers whose sum of digits equals a value How do I find all of the numbers in a given range whose sum of digits equal to a given value?
For example:
Range : 100 - 9000
Value : 4
Result : 103, 112, 121, 130, 202, 211, 220, 301, 310, ..., 4000, ...
 A: Observe that any four digit number larger than $4000$ will have a digit sum greater than $4$.  Thus, the problem reduces to finding the number of three or four digit numbers with digit sum $4$.
For three digit numbers, let $x_2$ denote the hundreds digit, $x_1$ denote the tens digit, and $x_0$ denote the units digit.  Then the number of three digit numbers whose digit sum is $4$ is the number of solutions of the equation
$$x_2 + x_1 + x_0 = 4 \tag{1}$$ 
in the non-negative integers subject to the constraint that $x_2 \geq 1$.  Let $x_2' = x_2 - 1$.  Then $x_2'$ is a non-negative integer.  Substituting $x_2' + 1$ for $x_2$ in equation 1 yields
\begin{align*}
x_2' + 1 + x_1 + x_0 & = 4\\
x_2' + x_1 + x_0 & = 3 \tag{2}
\end{align*}
which is an equation in the non-negative integers.  A particular solution of equation 2 corresponds to placing two addition signs in a row of three ones.  For instance, 
$$1 + 1 1 +$$
corresponds to the solution $x_2' = 1$, $x_1 = 2$, $x_3 = 0$, while 
$$1 + 1 + 1$$
corresponds to the solution $x_2' = x_1 = x_0 = 1$.  Thus, the number of solutions of equation 2 is the number of ways two addition signs can be inserted into a row of three ones, which is 
$$\binom{3 + 2}{2} = \binom{5}{2}$$
since we must choose which two of the five symbols (three ones and two addition signs) will be addition signs. 
For four digit numbers, let $x_2$, $x_1$, and $x_0$ be defined as above.  Let $x_3$ denote the thousands digit.  Then the number of four digit numbers whose digit sum is $4$ is the number of solutions of the equation
$$x_3 + x_2 + x_1 + x_0 = 4 \tag{3}$$
in the non-negative integers subject to the constraint that $x_3 \geq 1$.  Let $x_3' = x_3 - 1$.  Then $x_3$ is a non-negative integer.  Substituting $x_3' + 1$ for $x_3$ in equation 3, then simplifying yields
$$x_3' + x_2 + x_1 + x_0 = 3 \tag{4}$$
Equation 4 is an equation in the non-negative integers.  The number of solutions of equation 4 is the number of ways of inserting three addition signs in a row of three ones, which is 
$$\binom{3 + 3}{3} = \binom{6}{3}$$
since we must choose which three of the six symbols (three ones and three additions signs) will be addition signs.
In total, there are 
$$\binom{5}{2} + \binom{6}{3}$$
numbers between $100$ and $9000$ with digit sum $4$.  
