To determine which numbers $n$ from the given choices have 13 as a primitive root modulo $n$, we need to check the following conditions:

1. $n$ should be a prime number.
2. The order of 13 modulo $n$ should be $n-1$. This means that 13 raised to all powers from 1 to $n-2$ should not be congruent to 1 modulo $n$, and $13^{n-1}$ should be congruent to 1 modulo $n$.

Let's analyze each choice:

1) $n = 47$

• $n-1 = 46$
• Check if the order of 13 modulo 47 is 46 by verifying if $13^k \mod 47$ for $k$ from 1 to 45 is not 1, and $13^{46} \equiv 1 \mod 47$.

2) $n = 49$

• $n = 49$ is not a prime number (49 = $7^2$), so 13 cannot be a primitive root modulo 49.

3) $n = 53$

• $n-1 = 52$
• Check if the order of 13 modulo 53 is 52.

4) $n = 59$

• $n-1 = 58$
• Check if the order of 13 modulo 59 is 58.

5) $n = 61$

• $n-1 = 60$
• Check if the order of 13 modulo 61 is 60.

6) $n = 67$

• $n-1 = 66$
• Check if the order of 13 modulo 67 is 66.

Now, let's check these conditions step-by-step.

Steps:

1. Calculate $13^{k} \mod n$ for each $k$ from 1 to $n-1$ for all the prime numbers.
2. Identify the smallest $k$ such that $13^{k} \equiv 1 \mod n$. This is the order of 13 modulo $n$.
3. If this order is equal to $n-1$, then 13 is a primitive root modulo $n$.

Let’s compute these values for the prime choices.The numbers from the given choices where 13 is a primitive root modulo $n$ are:

• $47$
• $59$
• $67$

Would you like more details on how these calculations were performed, or do you have any other questions?

Here are 5 related questions to expand on this topic:

1. What are the properties of primitive roots in number theory?
2. How can we find primitive roots for any prime number $p$?
3. Why is 13 not a primitive root modulo some other numbers like 49 or 53?
4. What is the significance of the order of a number modulo $n$?
5. How does the concept of primitive roots apply in cryptography?

Tip: When checking if a number $g$ is a primitive root modulo $p$, ensure $p$ is a prime and verify if the smallest $k$ such that $g^k \equiv 1 \mod p$ equals $p-1$.