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The problem asks to show that the system of congruences has no solution:

\[
\begin{aligned}
x &\equiv 5 \pmod{6}, \\
x &\equiv 7 \pmod{15}.
\end{aligned}
\]

### Step-by-Step Analysis:

To determine if a solution exists, we can use the **Chinese Remainder Theorem** (CRT). The CRT guarantees a solution exists if the moduli \(6\) and \(15\) are coprime, and the system of congruences is consistent. However, if the moduli are not coprime or the system is inconsistent, there will be no solution.

1. **Prime Factorization of Moduli:**
   \[
   6 = 2 \times 3, \quad 15 = 3 \times 5.
   \]
   The greatest common divisor of 6 and 15 is \(\gcd(6, 15) = 3\). Since the moduli are not coprime, we need to check for compatibility between the congruences.

2. **Checking Compatibility:**
   - From \(x \equiv 5 \pmod{6}\), we know:
     \[
     x = 6k + 5 \quad \text{for some integer } k.
     \]
   - Substitute this into the second congruence \(x \equiv 7 \pmod{15}\):
     \[
     6k + 5 \equiv 7 \pmod{15}.
     \]
     Simplifying this:
     \[
     6k \equiv 2 \pmod{15}.
     \]
     Now, we need to solve:
     \[
     6k \equiv 2 \pmod{15}.
     \]
   - Divide both sides by \(\gcd(6, 15) = 3\) to simplify:
     \[
     2k \equiv \frac{2}{3} \pmod{\frac{15}{3}}.
     \]
     This yields a contradiction since \(2k \) cannot be \(\frac{2}{3}\) modulo anything.

Thus, the system has no solution due to the inconsistency in the moduli and the conflicting congruences.

Would you like further clarification, or do you have any other questions? 

Here are 5 related questions:
1. What is the Chinese Remainder Theorem?
2. How can we check if two moduli are coprime?
3. What are the conditions for a system of congruences to have a solution?
4. How would the solution change if the moduli were coprime?
5. Can the method of solving linear Diophantine equations help in this case?

**Tip:** Always check the greatest common divisor (gcd) of moduli when dealing with systems of congruences to assess whether a solution might exist.

Show that the system below does not have a solution:
x ≡ 5 (mod 6)
x ≡ 7 (mod 15)

Learn how to prove that the system of congruences x ≡ 5 (mod 6) and x ≡ 7 (mod 15) has no solution using modular arithmetic and the Chinese Remainder Theorem. This problem involves key concepts in number theory and is suitable for undergraduate-level mathematics.

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