To solve this system of simultaneous congruences, we can apply the Chinese Remainder Theorem (CRT). This theorem provides a solution when the moduli are pairwise coprime, as they are here (3, 5, and 13 are pairwise coprime). Here’s the process:

Given: $x \equiv 2 \pmod{3}$ $x \equiv 4 \pmod{5}$ $x \equiv 6 \pmod{13}$

Step 1: Find the product of the moduli

The product of the moduli $N$ is: $N = 3 \times 5 \times 13 = 195$

Step 2: Compute each partial product $N_i = \frac{N}{n_i}$ for each congruence

1. For $x \equiv 2 \pmod{3}$, $N_1 = \frac{195}{3} = 65$.
2. For $x \equiv 4 \pmod{5}$, $N_2 = \frac{195}{5} = 39$.
3. For $x \equiv 6 \pmod{13}$, $N_3 = \frac{195}{13} = 15$.

Step 3: Find the modular inverse of each $N_i$ modulo $n_i$

We need to find integers $M_1, M_2,$ and $M_3$ such that:

1. $65M_1 \equiv 1 \pmod{3}$
2. $39M_2 \equiv 1 \pmod{5}$
3. $15M_3 \equiv 1 \pmod{13}$

Calculating each $M_i$:

1. For $65M_1 \equiv 1 \pmod{3}$: $65 \equiv 2 \pmod{3}$ We need $2M_1 \equiv 1 \pmod{3}$, which gives $M_1 = 2$.

2. For $39M_2 \equiv 1 \pmod{5}$: $39 \equiv 4 \pmod{5}$ We need $4M_2 \equiv 1 \pmod{5}$, which gives $M_2 = 4$.

3. For $15M_3 \equiv 1 \pmod{13}$: $15 \equiv 2 \pmod{13}$ We need $2M_3 \equiv 1 \pmod{13}$, which gives $M_3 = 7$.

Step 4: Combine everything to find $x$

Now we can write the solution as: $x \equiv a_1 \cdot N_1 \cdot M_1 + a_2 \cdot N_2 \cdot M_2 + a_3 \cdot N_3 \cdot M_3 \pmod{N}$ where $a_1 = 2$, $a_2 = 4$, and $a_3 = 6$.

Substitute the values: $x \equiv 2 \cdot 65 \cdot 2 + 4 \cdot 39 \cdot 4 + 6 \cdot 15 \cdot 7 \pmod{195}$

Calculate each term:

1. $2 \cdot 65 \cdot 2 = 260$
2. $4 \cdot 39 \cdot 4 = 624$
3. $6 \cdot 15 \cdot 7 = 630$

Now, add these values together: $x \equiv 260 + 624 + 630 = 1514 \pmod{195}$

Finally, reduce $1514$ modulo $195$: $1514 \div 195 = 7 \quad \text{remainder } 149$ So, $x \equiv 149 \pmod{195}$

Answer

The solution to the system is: $x \equiv 149 \pmod{195}$

Would you like a breakdown of any specific steps or have further questions?

Related questions:

1. What is the Chinese Remainder Theorem and how does it work for systems with pairwise coprime moduli?
2. How can modular inverses be calculated for any congruence system?
3. What happens if the moduli in a system of congruences are not pairwise coprime?
4. How does one verify a solution obtained using the Chinese Remainder Theorem?
5. Can this method be used for systems with more than three congruences?

Tip: To simplify calculations with large numbers in modular arithmetic, breaking down each term step-by-step can reduce errors.