Given that 20251034B15A is a 12-digit number divisible by 21, find the minimum value of AB 

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

Given that 20251034B15A is a 12-digit number divisible by 21, find the minimum value of AB.

Solve the problem where the 12-digit number 20251034B15A is divisible by 21. This involves using the divisibility rules for 3 and 7 and modular arithmetic to determine the minimum value of AB. Suitable for high school students (Grades 9-12).

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