We are given that the 10-digit number $2020841A2B$ is divisible by 36, and $A > B > 0$. To solve for $A$, let's break down the requirements.

Divisibility by 36

A number divisible by 36 must be divisible by both 4 and 9:

1. Divisibility by 4: The last two digits of the number (i.e., $2B$) must form a number divisible by 4.
2. Divisibility by 9: The sum of the digits of the number must be divisible by 9.

Let's analyze these conditions:

Step 1: Divisibility by 4

The last two digits are $2B$. For $2B$ to be divisible by 4, the possible values for $B$ are: $2B \equiv 00, 04, 08, 12, 16, 20 \ (\text{mod}\ 4)$ Thus, $B = 0, 2, 4, 6, 8$.

Step 2: Divisibility by 9

The sum of the digits of the number $2020841A2B$ is: $2 + 0 + 2 + 0 + 8 + 4 + 1 + A + 2 + B = 19 + A + B$ For divisibility by 9, we need: $19 + A + B \equiv 0 \ (\text{mod}\ 9)$ This means: $19 + A + B = 27, 36, 45, \dots$ To satisfy both conditions, we will substitute possible values for $B$ and check which values of $A$ meet the conditions.

Step 3: Checking Possible Values for $A$ and $B$

Now, we will test the possibilities:

Case 1: $B = 2$

• Last two digits: $22$, divisible by 4.
• Sum: $19 + A + 2 = 21 + A$.
  • $21 + A = 27$ gives $A = 6$.

This satisfies $A > B > 0$ with $A = 6$ and $B = 2$.

Thus, the value of $A$ is 6.

Would you like more details on any step or have any questions?

Here are 5 related questions for further exploration:

1. What are the divisibility rules for other numbers like 11 or 7?
2. Can a different approach solve divisibility problems faster?
3. How do conditions like $A > B$ affect divisibility checks in general?
4. What if the number were divisible by 72 instead of 36?
5. How do we check divisibility of numbers in larger base systems?

Tip: Always start by breaking down divisibility problems using prime factorizations, as it helps in applying multiple rules systematically.