How many 4 digit numbers are divisible by 29 such that their digit sum is also 29? How many $4$ digit numbers are divisible by $29$ such that their digit sum is also $29$?
Well, answer is $5$ but what is the working and how did they get it?
 A: Let the digits be$b,c,d,29-b-c-d$
As $0\le b,c,d\le9,0<b+c+d\le27<29$
Now, $$(29-b-c-d)+10b+100c+1000d=29+9b+99c+999d\equiv-20b+12c-16d$$
We need  $29\mid(5b-3c+4d)\equiv-24b-3c+33d\iff29\mid(c+8b-11d)$
For $29\mid(8b-11d)\implies8b-11d=(29k-c)(11\cdot3-8\cdot4)$
$\iff8(b+116k-4c)=11(87k+d-3c)\iff b+116k=11m+4c$
$\iff b\equiv5k+4c\pmod{11}$ and $d\equiv3c+k\pmod{11}$
Test for $c;0\le c\le9$ ensuring $0\le b,c,d\le9$ and $b+c+d\ge20$
A: Here is a full solution of the problem: 
Let
$$(a,b,c,29-a-b-c)$$
be the decimal representation of such a number, and write $\equiv$ for equivalence modulo the prime $29$. We want $999a+99b+9c\equiv0$, or
$$13 a+12b+9c\equiv0\ .\tag{1}$$
Put $4b+3c=:r$. Then $(1)$ implies $13a+3r\equiv0$, which is satisfied by $a=2$, $r=1$, hence enforces
$$a\equiv2r\equiv 8b+6c\ .\tag{2}$$
We now have to determine the solutions of $(2)$ that in addition satisfy the constraints
$$a\in[1\>..\>9],\quad b,c\in[0\>..\>9],\quad 20\leq a+b+c\leq27\ .\tag{3}$$
This implies $b+c\geq11$, hence $b\geq2$, so that we can conclude that
$$70\leq 8b+6c\leq 126\ .$$
There are only the multiples $3\cdot29=87$ and $4\cdot 29=116$ near this range. In view of $(2)$ we therefore have to find the pairs $(b,c)$ with
$$88\leq 8b+6c\leq 96, \quad{\rm resp.,}\quad 118\leq 8b+6c\leq124\tag{4}$$ that lead to an $a$ such that the last constraint $(3)$ is satisfied. The first of the ranges $(4)$ contains the pairs
$$(9,4), \  (9,3), \ (8,5), \ (8,4), \ (7,6), \ (6,8), \ (6,7), \ (5,9), \ (5,8)\ ,$$
and the second range contains the pairs
$$(9,8), \ (8,9)\ .$$
For each of these pairs $(b,c)$ we now compute $a$ by means of $(2)$, and retain the triples $(a,b,c)$ for which the last condition $(3)$ is satisfied. These are the triples
$$(9,9,4), \ (7,8,5), \ (9,6,8), \ (7,5,9), \ (4,9,8)\ .$$
From these triples we then obtain the five numbers
$$9947, \ 7859, \ 9686, \ 7598, \ 4988$$
fulfilling the given conditions.
A: Just to give a different take on things, in order for a number to be a multiple of $29$ and have digit sum congruent to $2$ mod $9$, the number must be of the form 
$$29(9k+1)=261k+29$$
For the number to have four digits, we need $1000\le261k+29\le9999$, which requires $4\le k\le38$.  
Now for the digit sum to be not just congruent to $2$ mod $9$ but actually equal to $29$, we cannot have any $0$'s or $1$'s among the four digits, and if there's a $2$ then the other three digits must be $9$'s. Therefore, we cannot have $k\equiv1$ or $2$ mod $10$, nor, since $261$ does not divide $9992-29$, can we have $k\equiv3$ mod $10$; also,  we need $2999\le261k+29$, which by itself requires $12\le k$. At this point we're left with $19$ possible values of $k$:
$$14,15,16,17,18,19,20\\24,25,26,27,28,29,30\\34,35,36,37,38$$
The values that work turn out to be $k=19$, $29$, $30$, $37$, and $38$. These correspond to the numbers $4988$, $7598$, $7859$, $9686$, and $9947$ found by Christian Blatter, whose approach boiled things down to checking just $11$ possibilities instead of the $19$ listed here. I wonder if some hybrid of the two approaches might improve at least on mine if not both.
Added later: It occurs to me it's possible to winnow the list of $k$'s by about a third without too much effort.  If $k$ ends in $6$ or less, then $261k+29$ ends in $5$ or less, which means the first three digits must sum to at least $24$. The smallest such number is $6990$, from which it follows that $k\ge27$. This eliminates $k=14$, $15$, $16$, $24$, $25$, and $26$. This leaves only $13$ values of $k$ to check, which is closer to Christian's tally.
A: Given the $4$-digit number $\overline{abcd}$, the sum of digits being divisible by $29$ implies:
$$a+b+c+d=29,$$
because: $29<4\cdot 9<58$.
$\overline{abcd}$ is being divisible by $29$ implies:
$$1000a+100b+10c+d\equiv 14a+13b+10c+d\equiv 0\pmod{29}.$$
Hence:
$$\begin{cases} \ \ \ \ a+\ \ \ \ b+\ \ \ \ c+d=29 \\
14a+13b+10c+d=29n\end{cases}.$$
Multiply the first by $14$ and subtract the second:
$$b+4c+13d=29(14-n),9\le n<14,$$
because: $9+4\cdot 9+13\cdot 9=162<29\cdot 6=174$.
It is easy to check ($2\le b,c,d\le 9$):
$$\begin{align}&n=9 \Rightarrow b+4c+13d=145 \\
& \ \ \ d=9 \Rightarrow (b,c)=\color{red}{(4,6)}; (8,5);\\
&\ \ \ d=8 \Rightarrow (b,c)=(5,9); (9,8); \\
&\ \ \ (b,c,d)=(8,5,9), (5,9,8), (9,8,8);\\
&n=10 \Rightarrow b+4c+13d=116 \\
&\ \ \ d=8 \Rightarrow  (b,c)=\color{red}{(4,2)};\\
&\ \ \ d=7 \Rightarrow (b,c)=\color{red}{(5,5)}; (9,4);\\
&\ \ \ d=6 \Rightarrow (b,c)=\color{red}{(2,9)}; (6,8);\\
&\ \ \ (b,c,d)=(9,4,7), (6,8,6);\\
&n=11 \Rightarrow b+4c+13d=87 \\
&\ \ \ d=5 \Rightarrow  (b,c)=\color{red}{(2,5)}; \color{red}{(6,4)};\\
&\ \ \ d=4 \Rightarrow (b,c)=\color{red}{(3,8)}; \color{red}{(5,7)}; \color{red}{(9,6)}.
\end{align}$$
Hence:
$$\overline{abcd}=7859; 7598; 4988; 9947; 9686.$$
