Question

Guess me:

Make as many $3$-digit numbers as possible whose sum of the digits is $24$ (digits may be repeated).

Answer 1

Given number: $24$

As

$$\begin{gathered} 6+9+9=24 \\ 9+6+9=24 \\ 9+9+6=24 \\ 7+8+9=24 \\ 7+9+8=24 \\ 8+7+9=24 \\ 8+9+7=24 \\ 9+7+8=24 \\ 9+8+7=24 \\ 8+8+8=24 \end{gathered}$$

Hence, all possible $3$-digit numbers whose sum of the digits is $24$ are $699,969,996,789,798,879,897,978,987$ and $888$.