How many 4-digit positive integers divisible by 3 can be formed using only the digits \(\{1, 3, 4, 6, 7\}\), such that no digit appears more than once in a number?
Answer
(B)
Explanation
A number is divisible by 3 exactly when its digit sum is. The five digits add up to \(1 + 3 + 4 + 6 + 7 = 21\), and a 4-digit number with distinct digits from this set leaves out exactly one digit, so its digit sum is 21 minus the digit left out.
That is a multiple of 3 only when the digit left out is a multiple of 3, i.e. 3 or 6. Each choice leaves four distinct non-zero digits, which can be arranged in \(4! = 24\) ways.
Total \(= 2 \times 24 = 48\).
Answer: B