An 8-digit number 4252746B leaves remainder 0 when divided by 3. How many values of B are possible?

The correct option is 4.

To determine how many values of the single digit $B$ are possible such that the 8-digit number $4252746B$ is divisible by 3 (leaves a remainder of 0), we can apply the divisibility rule for 3.

The divisibility rule for 3 states that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

Let us find the sum of the digits of the given number:
$\text{Sum of digits}=4+2+5+2+7+4+6+B$
Simplifying this expression, we get:
$\text{Sum of digits}=30+B$

For the number $4252746B$ to leave a remainder of 0 when divided by 3, the sum of its digits must be a multiple of 3. Therefore:
$(30+B)$ must be divisible by 3.

Since 30 is already a multiple of 3, the value of $B$ itself must be divisible by 3. Since $B$ represents the units digit of the number, it must be a single-digit integer from the set of possible digits: $\{0,1,2,3,4,5,6,7,8,9\}$.

The single-digit values that are multiples of 3 (and thus divisible by 3) are:
$B=0,3,6,\text{or}9$

Counting these values, we find there are exactly 4 possible values for $B$ (which are 0, 3, 6, and 9) that make the 8-digit number divisible by 3.