To explain why the answer is $A = 4$, let's break this problem down step by step:

Problem Setup:

The given number is $AA77$, where $A$ is a digit, and the entire number must be divisible by $37$.

This can be written as: $AA77 = 1100A + 10A + 77 = 110A + 77$

We are tasked with finding a value of $A$ such that $110A + 77$ is divisible by $37$.

Divisibility Condition:

A number $N$ is divisible by $37$ if: $N \mod 37 = 0$

Thus, we need: $(110A + 77) \mod 37 = 0$

Simplifying Modulo 37:

1. Compute $110 \mod 37$:

   • Divide $110$ by $37$: $110 \div 37 = 2$ remainder $36$.
   • Therefore, $110 \mod 37 = 36$.

   So, the expression becomes: $(36A + 77) \mod 37 = 0$

2. Simplify $77 \mod 37$:

   • Divide $77$ by $37$: $77 \div 37 = 2$ remainder $3$.
   • Thus, $77 \mod 37 = 3$.

   The expression now becomes: $(36A + 3) \mod 37 = 0$

Solving for $A$:

The equation $(36A + 3) \mod 37 = 0$ implies: $36A + 3 = k \cdot 37 \quad \text{for some integer } k$

Rewriting: $36A \equiv -3 \pmod{37}$

Since $-3$ is equivalent to $34$ modulo $37$ (because $-3 + 37 = 34$), we have: $36A \equiv 34 \pmod{37}$

3. Solve for $A$ using the modular inverse of $36 \mod 37$:
   • The modular inverse of $36$ modulo $37$ is $1$ (since $36 \cdot 1 \equiv 1 \pmod{37}$).
   • Multiply both sides of $36A \equiv 34 \pmod{37}$ by $1$: $A \equiv 34 \pmod{37}$

Thus, $A = 34 \mod 10$ (because $A$ is a digit), which gives $A = 4$.

Final Answer:

The value of $A$ is $4$. When $A = 4$, the number $4477$ is divisible by $37$.

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Let me know if you'd like further details or have any questions. Here are 5 related questions for practice:

1. What is the modular inverse, and how is it calculated?
2. How can you test if a number is divisible by $37$ directly?
3. How does modular arithmetic simplify divisibility problems?
4. What happens if $A$ is replaced by a different digit? Can you verify divisibility manually?
5. What is the general process for solving divisibility problems with unknown digits?

Tip: Modular arithmetic is a powerful tool for solving divisibility problems efficiently. Practice breaking down