The 7-digit number 57A001B is divisible by 9. What is the minimum value of (A + B)?

The correct option is 5.

Let's verify and understand why this is the correct answer step-by-step.

A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

The given 7-digit number is 57A001B. Let's write down the sum of its digits:

$$\text{Sum of digits}=5+7+A+0+0+1+B$$

Simplifying the sum of the known digits:

$$5+7+0+0+1=13$$

Therefore, the sum of all digits is:

$$\text{Sum}=13+(A+B)$$

For the number to be divisible by 9, this sum must be a multiple of 9. That is:

$$13+(A+B)=9k$$

where $k$ is an integer.
Since A and B are single-digit non-negative integers (from 0 to 9), their sum $(A+B)$ must satisfy:
$$0\le A+B\le 18$$

This means the total sum $13+(A+B)$ must lie in the range:

$$13\le 13+(A+B)\le 31$$

The multiples of 9 that lie in this range are 18 and 27.

To find the minimum value of $(A+B)$, we equate the sum to the smallest possible multiple in this range, which is 18:

$$13+(A+B)=18$$

Solving for $(A+B)$:

$$A+B=18-13$$

$$A+B=5$$

Thus, the minimum value of $(A+B)$ is indeed 5.