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 ..

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