Solve the following equation using CRT.
x ≡ 3 (mod 6) x ≡ 6 (mod 7) x ≡ 10 (mod 11)

The given system of congruences is:

\[
\begin{aligned}
x &\equiv 3 \pmod{6} \\
x &\equiv 6 \pmod{7} \\
x &\equiv 10 \pmod{11}
\end{aligned}
\]

To solve this system using the **Chinese Remainder Theorem (CRT)**, we follow these steps:

### Step 1: Verify moduli are pairwise coprime
The CRT applies directly if the moduli are pairwise coprime. The moduli are \(6\), \(7\), and \(11\), and we need to check their greatest common divisors:
- \( \gcd(6, 7) = 1 \)
- \( \gcd(6, 11) = 1 \)
- \( \gcd(7, 11) = 1 \)

Since all the moduli are pairwise coprime, we can apply the Chinese Remainder Theorem.

### Step 2: Solve the system of congruences

Let the moduli be:
\[
n_1 = 6, \quad n_2 = 7, \quad n_3 = 11
\]
The solution \(x\) must satisfy:

\[
x \equiv 3 \pmod{6}, \quad x \equiv 6 \pmod{7}, \quad x \equiv 10 \pmod{11}
\]

The general form of the solution is:
\[
x = a_1 N_1 y_1 + a_2 N_2 y_2 + a_3 N_3 y_3 \pmod{N}
\]
where \(N = n_1 n_2 n_3\) and \(N_i = \frac{N}{n_i}\). The \(y_i\) values are the modular inverses of \(N_i\) modulo \(n_i\).

#### Step 2.1: Compute \(N\)
\[
N = 6 \times 7 \times 11 = 462
\]

#### Step 2.2: Compute \(N_i\) values
\[
N_1 = \frac{462}{6} = 77, \quad N_2 = \frac{462}{7} = 66, \quad N_3 = \frac{462}{11} = 42
\]

#### Step 2.3: Find the modular inverses
We need the inverses of \(N_1 = 77\) modulo \(6\), \(N_2 = 66\) modulo \(7\), and \(N_3 = 42\) modulo \(11\).

1. \(y_1\): Solve \(77y_1 \equiv 1 \pmod{6}\). We reduce \(77 \mod 6 = 5\), so \(5y_1 \equiv 1 \pmod{6}\). The inverse of 5 mod 6 is 5, so \(y_1 = 5\).

2. \(y_2\): Solve \(66y_2 \equiv 1 \pmod{7}\). We reduce \(66 \mod 7 = 3\), so \(3y_2 \equiv 1 \pmod{7}\). The inverse of 3 mod 7 is 5, so \(y_2 = 5\).

3. \(y_3\): Solve \(42y_3 \equiv 1 \pmod{11}\). We reduce \(42 \mod 11 = 9\), so \(9y_3 \equiv 1 \pmod{11}\). The inverse of 9 mod 11 is 5, so \(y_3 = 5\).

#### Step 2.4: Compute the solution
Now, substitute the values into the general solution formula:
\[
x = 3 \times 77 \times 5 + 6 \times 66 \times 5 + 10 \times 42 \times 5 \pmod{462}
\]

Simplifying each term:
\[
x = 3 \times 385 + 6 \times 330 + 10 \times 210
\]
\[
x = 1155 + 1980 + 2100 = 5235
\]

Finally, reduce modulo 462:
\[
x = 5235 \mod 462 = 165
\]

Thus, the solution is:

\[
x \equiv 165 \pmod{462}
\]

### Final Answer:
The solution to the system is \(x = 165 \pmod{462}\).

Would you like further clarification or details on any steps?

Here are 5 related questions to explore:
1. How do we compute the modular inverse for large numbers?
2. Can you apply the CRT when moduli are not pairwise coprime?
3. How does CRT work in practical applications like cryptography?
4. What if one of the moduli is not coprime to another—how would you solve it?
5. What is the difference between solving congruences using CRT versus direct substitution?

**Tip**: Always double-check your modular inverses, as errors here can lead to incorrect solutions!

Solve the following equation using CRT: x ≡ 3 (mod 6), x ≡ 6 (mod 7), x ≡ 10 (mod 11)

Learn how to solve a system of congruences using the Chinese Remainder Theorem (CRT). The problem involves solving x ≡ 3 (mod 6), x ≡ 6 (mod 7), and x ≡ 10 (mod 11). This tutorial covers key concepts in number theory, modular arithmetic, and the CRT theorem, suitable for Grades 11-12 and undergraduate students.

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