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?

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.
