How many 4-digit numbers can be formed using digits $1, 2, \dots , 9$ (without repetition) such that the sum of digits is divisible by 3?
Step-by-Step Solution
Key Concept: Digits 1–9 have three each of residues $0, 1, 2 \pmod 3$ ($\(\{3, 6, 9\}\), \(\{1, 4, 7\}\), \(\{2, 5, 8\}\)$). Enumerate selections $(a, b, c)$ (from each residue class) with $a + b + c = 4$ and $0a + 1b + 2c \equiv 0 \pmod 3$: valid triples are $(1, 0, 3), (2, 1, 1), (0, 2, 2), (1, 3, 0)$. Multiply ways to choose $\times 4!$ for each. Total $= 42 \times 4! = 1008$.
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Correct Answer: (1)