What is the smallest multiple of 29 whose last two digits in base 10 are 29 and whose sum of digits is 29?
A multiple of this form is $(100 * m + 1) \cdot 29$. The sum of the digits of $29m$ must therefore be 18. It must have at least three digits since $29 \nmid 99$, and $9 \mid m$. So we look at the successive multiples of $9 \cdot 29 = 261$. We find 783.