To solve for the minimum value of $A$ and $B$ such that the 12-digit number $20251034B15A$ is divisible by 21, we need to recall that a number divisible by 21 must be divisible by both 3 and 7.

Step 1: Divisibility by 3

For a number to be divisible by 3, the sum of its digits must be divisible by 3. Let’s calculate the sum of the known digits:

$2 + 0 + 2 + 5 + 1 + 0 + 3 + 4 + B + 1 + 5 + A = 23 + A + B$

We need $23 + A + B$ to be divisible by 3. In other words, we require:

$23 + A + B \equiv 0 \pmod{3}$

Thus,

$A + B \equiv 1 \pmod{3}$

Step 2: Divisibility by 7

Now, we use the divisibility rule for 7. One of the methods is to use modular arithmetic. We take the number $20251034B15A$ and express it mod 7. We use positional weights modulo 7, as follows:

The weights of the digits (from right to left) mod 7 are: $1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5$

So, the weighted sum is:

$5A \cdot 2 + 1 \cdot 5 + 1 \cdot B + 5 \cdot 4 + 1 \cdot 6 + 3 \cdot 3 + 0 \cdot 2 + 1 \cdot 5 + 2 \cdot 0 + 0 \cdot 1 + 2 \cdot 2$

This gives us a