Question

All possible 3-digit numbers are formed using the digits \(0,2,3,5,7,9\) without repeating any digit. Then the number of numbers among them which are divisible by \(15\) is

• A. \(11\)
• B. \(14\)
• C. \(16\)
• D. \(36\) \bigskip

Hint:

For divisibility by \(15\), first fix the last digit (\(0\) or \(5\)), then use the divisibility rule for \(3\). Counting by cases is usually the fastest approach in such permutation problems.

Answer 1

The Correct Option is A

Concept: Count numbers separately according to the last digit because divisibility by \(15\) requires divisibility by both \(3\) and \(5\).

Step 1: If the last digit is \(0\), the valid pairs are \((3,9),(2,7),(5,7)\), giving \(3\times2=6\) numbers.

Step 2: If the last digit is \(5\), the valid arrangements are \(705,375,735,795,975\), giving \(5\) numbers.

Step 3: Therefore, the total number of required three-digit numbers is \(6+5=11\). Hence, \(\boxed{11}\).