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

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Solution

Given number: $24$
As
$6 + 9 + 9 = 24 9 + 6 + 9 = 24 9 + 9 + 6 = 24 7 + 8 + 9 = 24 7 + 9 + 8 = 24 8 + 7 + 9 = 24 8 + 9 + 7 = 24 9 + 7 + 8 = 24 9 + 8 + 7 = 24 8 + 8 + 8 = 24$
Hence, all possible $3$ -digit numbers whose sum of the digits is $24$ are $699, 969, 996, 789, 798, 879, 897, 978, 987$ and $888$ .

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