GATE 2024 CY – 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 divisible by 3.
**Step 1: choose the digits.** A 4-digit number uses 4 of the 5 digits $\{1,3,4,6,7\}$, so it leaves out exactly one digit $d$. The total of all five digits is
$$1+3+4+6+7=21 .$$
The sum of the four digits used is $21-d$. Since 21 is a multiple of 3, $21-d$ is a multiple of 3 only if $d$ is a multiple of 3.
**Step 2: possible left-out digits.** Among the digits, the multiples of 3 are **3 and 6**, so $d=3$ or $d=6$.
**Step 3: arrangements.** For each choice the four remaining digits can be arranged in $4!=24$ ways.
$$2\times24=\mathbf{48}.$$