How many four digit numbers divisible by 29 have the sum of their digits 29? 
How many four digit numbers divisible by $29$ have the sum of their digits $29$?

A way to do it would be to write $1000a+100b+10c+d=29m$ and $a+b+c+d=29$ and then form equations like $14a+13b+10c+d=29m'$ and eventually $4a + 3b – 9d = 29 (m'' – 9)$. Analysing this equation for integer solutions using the advantage we have $\to$ $29$ is a prime; will give the solutions, but is tedious work.
Are there better solutions?
 A: If the sum of the digits of $n$ is $29$, then the number $n$ must be congruent to $2$ (modulo $9$). Since $29$ and $9$ are relatively prime, and their product is $261$, we need only consider numbers congruent to $29$ (modulo $261$). There are only about three dozen candidates between $1000$ and $9999$.
Furthermore, the average of the digits is $\frac{29}4>7$; no digit can be $1$, and only one digit can be below $6$. That means we can start our search at $2999$; the first candidate is $3161$, easily discarded, and we just repeatedly add $261$ until we get above $9999$.
A: $4 \times 9 = 36$, and we must end up with $29$ as a sum.
There are $7 \choose 4$ ways to reduce the sum from 36 to 29 over 4 digits.
Tedious = 35 cases.
A: $(a=4\land b=9\land c=8\land d=8\land m=172)\lor (a=7\land b=5\land c=9\land
   d=8\land m=262)\lor (a=7\land b=8\land c=5\land d=9\land m=271)\lor (a=9\land
   b=6\land c=8\land d=6\land m=334)\lor (a=9\land b=9\land c=4\land d=7\land
   m=343)$ 
So 5 numbers which are $4988,7598,7859,9686,9947$.
