JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is :
- (a) 36
- (b) 48
- (c) 60
- (d) 72
Answer and solution
Answer (d) 72
Solution
Maths sir · 3 steps
A multiple of 6 needs two things: digit sum divisible by 3 and an even last digit. Check them in that order.
- Divisible by 3. All six digits add to 1+2+3+5+6+7 = 24. A 5-digit number leaves out exactly one digit, so its digit sum is 24 minus that digit. It is divisible by 3 only when the left-out digit is 3 or 6.
- Divisible by 2. The units digit must be even. Fix it first, then arrange the other 4 digits in 4! = 24 ways.
Digit left out Digits used Even choices for units place Numbers 3 1, 2, 5, 6, 7 (sum 21) 2 or 6, so 2 ways 2 × 4! = 48 6 1, 2, 3, 5, 7 (sum 18) only 2, so 1 way 1 × 4! = 24 - So. The total is 48 + 24 = 72, which is option (d).
The idea to keep. For divisibility by 6, first use the digit sum to decide which digits are allowed, then fix an even digit in the units place and arrange the rest.
Answer from the NTA final answer key. Solution by PrepWiser.