Discussion
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
There are 48 such numbers.
A number is divisible by 3 when its digit sum is. The five digits sum to 1 + 3 + 4 + 6 + 7 = 21. Choosing four of them means leaving one out, and the remaining sum, 21 minus that digit, is divisible by 3 only if the omitted digit is itself divisible by 3, that is 3 or 6.
So the usable digit sets are {1, 4, 6, 7} and {1, 3, 4, 7}. Each can be arranged in 4! = 24 ways, giving 2 × 24 = 48. 24 counts only one of the two sets, and 72 or 12 do not follow from the count.
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