GATE 2024 DA – Question 3
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?
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Correct answer: (B) 48
Explanation
A number is divisible by 3 when the sum of its digits is. The sum of all five digits is $1 + 3 + 4 + 6 + 7 = 21$. A 4-digit number leaves out one digit, and the remaining sum $21 - d$ is divisible by 3 only if the left-out digit $d$ is divisible by 3, that is 3 or 6. Each of these two choices gives $4! = 24$ arrangements, so there are $2 \times 24 = 48$ numbers.