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We are now looking at congruences rather than equations. So we want the solution to be in the form of a congruence as well. For example, for 8(b), your solution for x should be of the form x ≡ a(mod 50). You can double check that your solution is correct by calculating 13(a) mod 50 and it should be 42.

Let's solve each linear congruence in the required form.

### Part (a): \( 15x \equiv 9 \pmod{25} \)

To solve this, we need to find all values of \( x \) that satisfy:
\[
15x \equiv 9 \pmod{25}.
\]

1. First, let's check if there is a solution by examining the gcd of 15 and 25.
   - \( \gcd(15, 25) = 5 \).
   - Since 5 divides the right-hand side (9), a solution exists.

2. Simplify the congruence by dividing everything by the gcd (5):
   \[
3x \equiv \frac{9}{5} \pmod{\frac{25}{5}}.
\]
   - Simplified equation:
   \[
3x \equiv 9 \pmod{5}.


For each of the following linear congruences, find all of its solutions.

(a) 15x ≡ 9 (mod 25)
(b) 13x ≡ 42 (mod 50)

Learn how to solve linear congruences such as 15x ≡ 9 (mod 25) and 13x ≡ 42 (mod 50) with detailed solutions. This guide explores key concepts in modular arithmetic and number theory, including divisibility and the greatest common divisor (gcd). Suitable for students in Grades 10-12.

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