How many 3-digit numbers divisible by 5 can be formed using the digits 0, 1, 4, 5, 6 and 8, if no digit is repeated in a number?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 36
A number divisible by 5 ends in 0 or 5. The two cases are counted separately, because 0 can never be the first digit and that changes the count.
Ends in 0: hundreds place any of the other 5 digits, tens place 4 left: \(5\times4=20\)
Ends in 5: hundreds place cannot be 0 or 5, so 4 choices (1, 4, 6, 8); tens place: 4 left (0 is now allowed): \(4\times4=16\)
Numbers \(=20+16=36\)
Why the other options are incorrect
Option (1) 40 — Incorrect. This is \(2\times20\); when the number ends in 5, the first digit cannot be 0, so that case gives only 16.
Option (3) 20 — Incorrect. This counts only the numbers ending in 0.
Option (4) 16 — Incorrect. This counts only the numbers ending in 5.