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Let's break down the questions one by one:

### a) Find \( d = \gcd(40, 900) \) and express it in the form \( d = s(40) + t(900) \) where \( s, t \in \mathbb{Z} \). Hence, find \( \text{LCM}(40, 900) \).

1. **Finding \( \gcd(40, 900) \)**:
   Using the Euclidean algorithm:
   \[
   900 = 22(40) + 20 \\
   40 = 2(20) + 0
   \]
   So, \( \gcd(40, 900) = 20 \).

2. **Expressing \( \gcd \) as a linear combination**:
   Working backwards:
   \[
   20 = 900 - 22(40)
   \]
   This gives \( s = -22 \) and \( t = 1 \), so:
   \[
   d = 1(900) - 22(40)
   \]

3. **Finding \( \text{LCM}(40, 900) \)**:
   The formula for LCM is:
   \[
   \text{LCM}(a, b) = \frac{a \cdot b}{\gcd(a, b)}
   \]
   Thus:
   \[
   \text{LCM}(40, 900) = \frac{40 \times 900}{20} = 1800
   \]

---

### b) Find \( d = \gcd(180, 168) \) and \( \text{LCM}(180, 168) \). Write \( d \) in the form \( s(180) + t(168) \).

1. **Finding \( \gcd(180, 168) \)**:
   Using the Euclidean algorithm:
   \[
   180 = 1(168) + 12 \\
   168 = 14(12) + 0
   \]
   So, \( \gcd(180, 168) = 12 \).

2. **Expressing \( \gcd \) as a linear combination**:
   Working backwards:
   \[
   12 = 180 - 1(168)
   \]
   This gives \( s = 1 \) and \( t = -1 \), so:
   \[
   d = 1(180) - 1(168)
   \]

3. **Finding \( \text{LCM}(180, 168) \)**:
   Using the LCM formula:
   \[
   \text{LCM}(180, 168) = \frac{180 \times 168}{12} = 2520
   \]

---

### c) Solve the following system of equations:

#### i) Solve the system:
   \[
   x \equiv 3 \pmod{7} \\
   x \equiv 6 \pmod{10}
   \]

   First, find a solution using the Chinese Remainder Theorem (CRT). We need to solve:
   \[
   x = 7k + 3 \quad \text{(from the first equation)}
   \]
   Substitute this into the second equation:
   \[
   7k + 3 \equiv 6 \pmod{10} \\
   7k \equiv 3 \pmod{10}
   \]
   Solve for \( k \) by trial or using modular arithmetic. The solution is \( k = 7 \), so:
   \[
   x = 7(7) + 3 = 52
   \]
   Therefore, \( x = 52 \pmod{70} \).

#### ii) Solve the system:
   \[
   x \equiv 5 \pmod{8} \\
   x \equiv 4 \pmod{11}
   \]
   Using the same method, set:
   \[
   x = 8k + 5
   \]
   Substitute into the second equation:
   \[
   8k + 5 \equiv 4 \pmod{11} \\
   8k \equiv -1 \equiv 10 \pmod{11}
   \]
   Multiply both sides by the inverse of 8 modulo 11, which is 7 (because \( 8 \times 7 \equiv 1 \pmod{11} \)):
   \[
   k \equiv 70 \pmod{11} \quad \text{or} \quad k \equiv 4 \pmod{11}
   \]
   So, \( k = 11m + 4 \). Substituting this back:
   \[
   x = 8(11m + 4) + 5 = 88m + 37
   \]
   Therefore, \( x = 37 \pmod{88} \).

---

Would you like further details or clarifications on any step?

### Follow-up Questions:
1. Can you explain how the Euclidean algorithm works in more detail?
2. What are the properties of the LCM and GCD that make them related?
3. How do you find the modular inverse, as used in part c?
4. Can the Chinese Remainder Theorem be used for more than two moduli?
5. What other methods exist for solving systems of congruences?

### Tip:
When working with modular arithmetic, it's useful to check if the moduli are coprime, as it simplifies the application of the Chinese Remainder Theorem.

a) Find GCD(40,900) and write d=s(40) + t(900), s, t ∈ Z. Hence find LCM(40,900). b) Find GCD(180,168) and LCM(180,168). Write d in the form of s(180) + t(168). c) Solve the following systems of equations: i) x ≡ 3 (mod 7) and x ≡ 6 (mod 10). ii) x ≡ 5 (mod 8) and x ≡ 4 (mod 11).

Learn how to find the GCD and LCM using the Euclidean algorithm and solve systems of congruences with the Chinese Remainder Theorem. This example includes detailed solutions for GCD(40,900), LCM(40,900), GCD(180,168), LCM(180,168), and solving modular arithmetic systems.

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