Solving Congruences

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🔮 The Magic of Clock Math: Solving Congruences

Imagine you have a magical clock that helps you solve secret puzzles. Let’s discover how!


🎯 What’s This All About?

You know how a clock goes 1, 2, 3… up to 12, and then starts over at 1 again? That’s the heart of modular arithmetic! Today, we’ll learn how to solve puzzles using this clock magic.

Our Big Idea: Finding a number that, when we put it through our clock, gives us exactly what we want.


📖 Linear Congruences: The Secret Messages

What is a Linear Congruence?

Think of it like a secret code:

“I’m thinking of a number. When I multiply it by 3 and look at it on a clock-7, I get 4. What’s my number?”

Written in math language:

3x ≡ 4 (mod 7)

Breaking it down:

  • 3x = 3 times some mystery number
  • = “is the same as on our clock”
  • 4 = what we want to get
  • (mod 7) = we’re using a clock that goes 0 to 6

🌟 Simple Example

Puzzle: Find x where 2x ≡ 6 (mod 8)

Think like this:

  • What number, times 2, gives something that shows 6 on a clock-8?
  • Try x = 3: 2 × 3 = 6 ✓
  • Try x = 7: 2 × 7 = 14 → on clock-8 = 6 ✓

Answer: x = 3 or x = 7 (and they repeat every 8!)


🔧 Solving Linear Congruences

The Key Rule

For ax ≡ b (mod n) to have a solution:

  • Find GCD(a, n) — the biggest number that divides both a and n
  • If GCD(a, n) divides b evenly, we have solutions!
  • If not, no solution exists.

🎮 Step-by-Step Method

Problem: Solve 5x ≡ 3 (mod 7)

graph TD A[Start: 5x ≡ 3 mod 7] --> B[Find GCD of 5 and 7] B --> C[GCD = 1] C --> D[1 divides 3? YES!] D --> E[Find inverse of 5 mod 7] E --> F[5 × 3 = 15 ≡ 1 mod 7] F --> G[Inverse of 5 is 3] G --> H[x = 3 × 3 = 9 ≡ 2 mod 7] H --> I[Answer: x = 2]

Let’s verify: 5 × 2 = 10 → on clock-7 = 3 ✓

📚 Another Example

Problem: Solve 4x ≡ 2 (mod 6)

Step 1: GCD(4, 6) = 2

Step 2: Does 2 divide 2? Yes! So solutions exist.

Step 3: Simplify by dividing everything by 2:

  • 4x ≡ 2 (mod 6) becomes 2x ≡ 1 (mod 3)

Step 4: Find x:

  • Try x = 2: 2 × 2 = 4 → on clock-3 = 1 ✓

Answer: x = 2, and also x = 5 (since 2 + 3 = 5)


🎪 Systems of Congruences: Multiple Clocks!

What if We Have Multiple Puzzles?

Imagine your friend says:

“I’m thinking of a number. On clock-3, it shows 2. On clock-5, it shows 3. What is it?”

This is a system of congruences:

x ≡ 2 (mod 3)
x ≡ 3 (mod 5)

🔍 Finding the Answer

Method: List and Match

Numbers that show 2 on clock-3:

2, 5, 8, 11, 14, 17, 20, 23…

Numbers that show 3 on clock-5:

3, 8, 13, 18, 23, 28…

The match: 8 appears in both lists!

So x = 8 is our answer!

And the pattern repeats every 3 × 5 = 15:

x = 8, 23, 38, 53…


👑 The Chinese Remainder Theorem (CRT)

The Ancient Secret

Over 1,500 years ago, a Chinese mathematician discovered something magical:

If your clocks have no common factors, there’s ALWAYS exactly one answer (before things repeat)!

🎯 The Rule

If GCD(m, n) = 1 (the clocks share no factors), then:

x ≡ a (mod m)
x ≡ b (mod n)

has exactly ONE solution between 0 and (m × n - 1).

🌈 CRT Example

Problem: Find x where:

  • x ≡ 1 (mod 3)
  • x ≡ 2 (mod 5)
graph TD A[x ≡ 1 mod 3] --> C[Numbers: 1, 4, 7, 10, 13...] B[x ≡ 2 mod 5] --> D[Numbers: 2, 7, 12, 17...] C --> E[Find common number] D --> E E --> F[x = 7!] F --> G[Pattern repeats every 15]

Check:

  • 7 on clock-3 = 1 ✓
  • 7 on clock-5 = 2 ✓

The Formula Way:

  1. M = 3 × 5 = 15
  2. M₁ = 15/3 = 5, M₂ = 15/5 = 3
  3. Find y₁ where 5y₁ ≡ 1 (mod 3) → y₁ = 2
  4. Find y₂ where 3y₂ ≡ 1 (mod 5) → y₂ = 2
  5. x = (1 × 5 × 2) + (2 × 3 × 2) = 10 + 12 = 22
  6. 22 mod 15 = 7

🎢 Real-World Magic

The Egg Problem:

A farmer has eggs. When counted by 3s, 2 remain. When counted by 5s, 3 remain. When counted by 7s, 2 remain. How many eggs?

x ≡ 2 (mod 3)
x ≡ 3 (mod 5)
x ≡ 2 (mod 7)

Using CRT step by step:

  1. Solve first two: x ≡ 8 (mod 15)
  2. Now solve: x ≡ 8 (mod 15) and x ≡ 2 (mod 7)
  3. Numbers ≡ 8 (mod 15): 8, 23, 38, 53…
  4. Check which ≡ 2 (mod 7): 23 ÷ 7 = 3 remainder 2 ✓

Answer: 23 eggs (and 128, 233, etc.)


🧩 Quick Summary

Concept What It Means Example
Linear Congruence ax ≡ b (mod n) 3x ≡ 6 (mod 9)
Solution Exists? GCD(a,n) must divide b GCD(3,9)=3, 3÷6? Yes!
System Multiple conditions x≡1(mod 3), x≡2(mod 5)
CRT If GCDs are 1, one unique solution exists Answer: x=7

💡 Pro Tips

🚀 Tip 1: Always check if a solution exists before solving!

🧙‍♂️ Tip 2: For CRT, list out a few values for each condition and look for matches.

⚠️ Tip 3: The answer repeats every (product of all moduli) steps.


🎉 You Did It!

You now understand:

  • ✅ What linear congruences are
  • ✅ How to solve them step by step
  • ✅ How systems of congruences work
  • ✅ The magical Chinese Remainder Theorem

Clock math isn’t just math—it’s like having a secret decoder ring for numbers! 🔐

Next time someone asks you a puzzle about remainders, you’ll know exactly what to do.

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