Question

The number of 3-digit integers that are multiple of 6 which can be formed by using the digits 1, 2, 3, 4, 5, 6 without repetition is:

• A. 22
• B. 26
• C. 20
• D. 24

Hint:

To determine the number of numbers divisible by 6, ensure the number meets both divisibility rules for 2 and 3.

Answer 1

The Correct Option is D

A number is divisible by 6 if it is divisible by both 2 and 3. - A number is divisible by 2 if its last digit is even. The available even digits are 2, 4, and 6. 

- A number is divisible by 3 if the sum of its digits is divisible by 3. 

Let's first calculate the total number of 3-digit integers that can be formed with the digits 1, 2, 3, 4, 5, 6 without repetition. The total number of such 3-digit numbers is: \[ 6 \times 5 \times 4 = 120 \] 

Now, we need to consider two conditions: the last digit must be even (for divisibility by 2), and the sum of the digits must be divisible by 3 (for divisibility by 3). 

For divisibility by 2, the last digit must be one of 2, 4, or 6. We can then count the number of possible 3-digit numbers for each case where the last digit is even, and the sum of the digits is divisible by 3. After checking all possibilities, we find that there are 24 valid numbers that meet both criteria.