Let $m$ and $n$ be non-negative integers. If $m = 6n + 2$, then what integer between $0$ and $m$ is the inverse of $3$ modulo $m$? Answer in terms of $n$.

We are tasked with finding an integer between $ 0 $ and $ m $ that is the inverse of $ 3 $ modulo $ m $, where $ m = 6n + 2 $, and $ n $ is a non-negative integer.

### Step 1: Set up the problem

We need to find an integer $ x $ such that:
\[
3x \equiv 1 \pmod{m}
\]
where $ m = 6n + 2 $. This equation means that $ 3x - 1 $ must be divisible by $ m $, i.e., there exists an integer $ k $ such that:
\[
3x = 1 + km
\]
or equivalently:
\[
3x \equiv 1 \pmod{6n+2}.
\]

### Step 2: Investigate values of $ m $

We substitute $ m = 6n + 2 $ into the equation $ 3x \equiv 1 \pmod{m} $, which gives us:
\[
3x \equiv 1 \pmod{6n+2}.
\]
Our goal is to find $ x $ that satisfies this congruence.

### Step 3: Check small values of $ n $

#### Case 1: $ n = 0 $
If $ n = 0 $, then $ m = 6(0) + 2 = 2 $. The equation becomes:
\[
3x \equiv 1 \pmod{2}.
\]
Since $ 3 \equiv 1 \pmod{2} $, the equation simplifies to:
\[
x \equiv 1 \pmod{2}.
\]
Thus, $ x = 1 $ is the solution for $ n = 0 $.

#### Case 2: $ n = 1 $
If $ n = 1 $, then $ m = 6(1) + 2 = 8 $. The equation becomes:
\[
3x \equiv 1 \pmod{8}.
\]
We can check small values of $ x $:
- For $ x = 1 $, $ 3(1) = 3 \not\equiv 1 \pmod{8} $.
- For $ x = 2 $, $ 3(2) = 6 \not\equiv 1 \pmod{8} $.
- For $ x = 3 $, $ 3(3) = 9 \equiv 1 \pmod{8} $.

Thus, $ x = 3 $ is the solution for $ n = 1 $.

#### Case 3: $ n = 2 $
If $ n = 2 $, then $ m = 6(2) + 2 = 14 $. The equation becomes:
\[
3x \equiv 1 \pmod{14}.
\]
We check small values of $ x $:
- For $ x = 1 $, $ 3(1) = 3 \not\equiv 1 \pmod{14} $.
- For $ x = 2 $, $ 3(2) = 6 \not\equiv 1 \pmod{14} $.
- For $ x = 3 $, $ 3(3) = 9 \not\equiv 1 \pmod{14} $.
- For $ x = 5 $, $ 3(5) = 15 \equiv 1 \pmod{14} $.

Thus, $ x = 5 $ is the solution for $ n = 2 $.

### Step 4: Generalize the solution

We observe a pattern in the solutions for $ x $:
- For $ n = 0 $, $ x = 1 $.
- For $ n = 1 $, $ x = 3 $.
- For $ n = 2 $, $ x = 5 $.

The values of $ x $ increase by 2 as $ n $ increases. Hence, the solution can be written as:
\[
x = 2n + 1.
\]

### Final Answer:
The integer between $ 0 $ and $ m $ that is the inverse of $ 3 $ modulo $ m $, where $ m = 6n + 2 $, is $ \boxed{2n + 1} $.

Would you like further details or explanations on any part of this solution?

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Here are 5 related questions for further exploration:
1. How would you solve $ 5x \equiv 1 \pmod{m} $ for arbitrary $ m $?
2. What is the general method to find the modular inverse of any integer?
3. Can you prove that $ 3x \equiv 1 \pmod{m} $ has a solution for all $ m $?
4. How does the pattern in the solution relate to properties of modular arithmetic?
5. What would the inverse of other integers modulo $ m = 6n + 2 $ look like?

**Tip**: When solving modular equations, checking patterns for small values often reveals a general formula that applies to larger cases!

Explore how to find the integer between 0 and m that is the inverse of 3 modulo m, where m = 6n + 2. This problem utilizes modular arithmetic and the modular inverse theorem. Perfect for students in Grades 10-12.

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