Question

Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2, 3, 4, 5 with repetition, is___.

Answer 1

Correct Answer: 16

For a number to be divisible by 6, it must be divisible by both 2 and 3.
Divisibility by 2: The last digit (unit place) must be an even number.
From the available digits 1, 2, 3, 4, 5, the even digits are 2 and 4.
Hence, the last digit must be either 2 or 4.
Divisibility by 3: The sum of the digits must be divisible by 3.
For the last digit being 2:
The first two digits (a, b) must be chosen such that the sum \( a + b + 2 \) is divisible by 3. The possible pairs for \( a + b = 1, 3, 5, 7, 9 \) (mod 3) are (1, 3), (3, 1), (2, 5), (5, 2), (4, 3), and (3, 4).
For the last digit being 4:
Similarly, \( a + b + 4 \) must be divisible by 3, yielding 8 possible combinations.
Thus, the total number of valid combinations is 8 + 8 = 16.