For a positive integer b > 1, if the product of two numbers 6344 and 42b8 is divisible by 12, then find the least value of b.

Correct option is D

Given:

Two numbers: 6344 and 42b8,

The product of these two numbers is divisible by 12.

Concept Used:

A number is divisible by 12 if it is divisible by both 3 and 4.

Divisibility rule for 3: The sum of the digits of the number must be divisible by 3.

Divisibility rule for 4: The last two digits of the number must be divisible by 4.

Solution:

Checking divisibility by 4:

For the product to be divisible by 4, at least one of the numbers (6344 or 42b8) must be divisible by 4.

The number 6344 ends with 44, and 44 ÷ 4 = 11, so 6344 is divisible by 4.

Therefore, the product is already divisible by 4, regardless of the value of b.

Checking divisibility by 3:

For the product to be divisible by 3, at least one of the numbers (6344 or 42b8) must be divisible by 3.

First, if 6344 is divisible by 3:

Sum of digits of 6344 = 6 + 3 + 4 + 4 = 17

Since 17 is not divisible by 3, 6344 is not divisible by 3.

Now, if 42b8 is divisible by 3:

Sum of digits of 42b8 = 4 + 2 + b + 8 = 14 + b

For 42b8 to be divisible by 3, 14 + b must be divisible by 3.

The smallest b>1that satisfies this condition is b = 4, because:

14 + 4 = 18, and 18 is divisible by 3.

Thus, least possible value of b is 4