For a number to be divisible by 28, it must be divisible by both 4 and 7.

Step 1: Check for divisibility by 4

A number is divisible by 4 if its last two digits form a number that is divisible by 4.

The last two digits here are B68, so we consider B6 as a two-digit number.

For B6 to be divisible by 4: $B6 \mod 4 = 0$ Possible values for B6: 16, 20, 24, 28, 32, 36...

Since $B > 0$, the smallest valid B is 2, making B6 = 26.

Step 2: Check for divisibility by 7

The full number is 13A2646268 (substituting $B = 2$).

A number is divisible by 7 if: $13A2646268 \mod 7 = 0$ Testing A = 1 first: $1312646268 \mod 7 = 0$ It is divisible by 7 when A = 1.

Step 3: Find $A + B$

$A = 1, B = 2$ $A + B = 1 + 2 = \mathbf{3}$

Thus, the smallest value of $A + B$ is 3.

---

Would you like a detailed explanation of how we checked divisibility by 7?

Related Questions:

1. How do you check divisibility by 7 using modular arithmetic?
2. Can a number be divisible by 28 but not by 4 or 7 separately?
3. What is the general rule for checking divisibility by 4?
4. How do you determine the smallest possible values in similar divisibility problems?
5. What are other methods to check divisibility for large numbers?

Math Tip:

For divisibility by 7, a quick test is to double the last digit, subtract from the rest, and check if the result is divisible by 7!