To explain the answer step by step:

Problem Statement Recap:

We are tasked with finding the integer $c$ such that: $c \equiv 13a \pmod{19}$ where $a \equiv 11 \pmod{19}$. The value of $c$ must satisfy $0 \leq c \leq 18$.

Step-by-Step Explanation:

1. Substitute $a$ into the equation: Since $a \equiv 11 \pmod{19}$, substitute $a = 11$ into the congruence: $c \equiv 13 \cdot 11 \pmod{19}$

2. Compute $13 \cdot 11$: Perform the multiplication: $13 \cdot 11 = 143$

3. Reduce $143 \mod 19$: To find the equivalent value of $143 \mod 19$, divide $143$ by $19$ and take the remainder: $143 \div 19 = 7 \text{ remainder } 10$ Thus: $143 \equiv 10 \pmod{19}$

4. Conclude $c$: The integer $c$ satisfying $c \equiv 13a \pmod{19}$ is: $c = 10$

Why This Works:

Modulo arithmetic ensures that we only care about the remainder when dividing by the modulus (19 in this case). By simplifying the large product $13 \cdot 11$ modulo 19, we find the equivalent smaller value, which is $10$.

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Would you like further details or clarification?

Here are 5 related questions to explore this topic:

1. What is the general procedure for solving modular arithmetic problems like this?
2. How would the solution change if $a \equiv 11$ and $b \equiv 3$ were under a different modulus, such as 17 or 23?
3. What is the significance of modular arithmetic in real-world applications?
4. Can we always reduce products mod $n$, and are there any special rules when the modulus is prime?
5. How can modular arithmetic be extended to solve systems of congruences?

Tip: Always simplify intermediate steps when working with large numbers in modular arithmetic to avoid computational errors!