How many six-digit positive numbers can be formed using the digits 1, 2, 3, 4, 5, and 6 exactly once, such that the six–digit number is divisible by its units digit, if the units digit is not 1?

Correct answer marked below

456

492

520

528

Correct

620

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Explanation

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Official Explanation

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Ending with 2 - All

5!

(120) possible numbers are divisible by 2.

Ending with 3 - All

5!

(120) possible numbers are divisible by 3.

Ending with 4 - the tens digit has to be 2 or 6 for the number to be divisible by 4, so only

2 \times 4!

numbers are divisible - 48 numbers.…

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What this question tests

Whether you can count arrangements and selections correctly — permutations vs. combinations.

Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.