Find all x ∈ Z satisfying each of the following equations.
(a) 3x ≡ 2 (mod 7)
(b) 5x + 1 ≡ 13 (mod 23)
(c) 5x + 1 ≡ 13 (mod 26)
(d) 9x ≡ 3 (mod 5)
(e) 5x ≡ 1 (mod 6)
(f) 3x ≡ 1 (mod 6)

Let's solve each congruence equation step by step, finding the solutions for \( x \in \mathbb{Z} \).

### (a) \( 3x \equiv 2 \pmod{7} \)
We need to solve \( 3x \equiv 2 \pmod{7} \). To do this, first, find the inverse of 3 modulo 7.

The inverse of 3 modulo 7 is the number \( y \) such that \( 3y \equiv 1 \pmod{7} \). We can check different values of \( y \) to find the inverse:

- \( 3 \times 1 = 3 \not\equiv 1 \pmod{7} \)
- \( 3 \times 2 = 6 \not\equiv 1 \pmod{7} \)
- \( 3 \times 3 = 9 \equiv 2 \pmod{7} \)
- \( 3 \times 4 = 12 \equiv 5 \pmod{7} \)
- \( 3 \times 5 = 15 \equiv 1 \pmod{7} \) (Found the inverse!)

So, the inverse of 3 modulo 7 is 5. Multiply both sides of the congruence by 5:
\[
5 \times 3x \equiv 5 \times 2 \pmod{7}
\]
\[
x \equiv 10 \pmod{7}
\]
Since \( 10 \equiv 3 \pmod{7} \), the solution is:
\[
x \equiv 3 \pmod{7}
\]

### (b) \( 5x + 1 \equiv 13 \pmod{23} \)
First, subtract 1 from both sides:
\[
5x \equiv 12 \pmod{23}
\]
Now, we need to find the inverse of 5 modulo 23. Testing different values of \( y \):

- \( 5 \times 1 = 5 \)
- \( 5 \times 2 = 10 \)
- \( 5 \times 3 = 15 \)
- \( 5 \times 4 = 20 \)
- \( 5 \times 5 = 25 \equiv 2 \pmod{23} \)
- \( 5 \times 9 = 45 \equiv 22 \pmod{23} \)
- \( 5 \times 14 = 70 \equiv 1 \pmod{23} \) (Found the inverse!)

So, the inverse of 5 modulo 23 is 14. Multiply both sides by 14:
\[
14 \times 5x \equiv 14 \times 12 \pmod{23}
\]
\[
x \equiv 168 \pmod{23}
\]
Since \( 168 \equiv 7 \pmod{23} \), the solution is:
\[
x \equiv 7 \pmod{23}
\]

### (c) \( 5x + 1 \equiv 13 \pmod{26} \)
Subtract 1 from both sides:
\[
5x \equiv 12 \pmod{26}
\]
Since 5 and 26 are not coprime (their gcd is 1), we need to check if this equation has a solution. To see if a solution exists, divide the whole equation by \( \gcd(5, 26) = 1 \):
\[
x \equiv \frac{12}{1} \pmod{\frac{26}{1}} \implies x \equiv ..
### Solution ` look

This page provides detailed solutions to various modular arithmetic equations, such as 3x ≡ 2 (mod 7) and 5x + 1 ≡ 13 (mod 23). The step-by-step breakdown covers key concepts like finding modular inverses and solving congruence equations. Suitable for students in Grades 11-12.

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