To find the order of the element $3$ in the multiplicative group $\mathbb{Z}_{31}^*$, we need to determine the smallest positive integer $k$ such that:

$3^k \equiv 1 \ (\text{mod} \ 31)$

Here's how we proceed:

1. Check the order of the group: The order of the multiplicative group $\mathbb{Z}_{31}^*$ is $\phi(31) = 31 - 1 = 30$, where $\phi$ is Euler's totient function. Therefore, the possible values of $k$ are divisors of 30.

2. Test each divisor: The divisors of 30 are $1, 2, 3, 5, 6, 10, 15, 30$. We need to find the smallest $k$ for which $3^k \equiv 1 \ (\text{mod} \ 31)$.

   • $3^1 \equiv 3 \ (\text{mod} \ 31)$
   • $3^2 = 9 \equiv 9 \ (\text{mod} \ 31)$
   • $3^3 = 27 \equiv 27 \ (\text{mod} \ 31)$
   • $3^5 = 243 \equiv 26 \ (\text{mod} \ 31)$
   • $3^6 = 3^3 \times 3^3 = 27 \times 27 = 729 \equiv 16 \ (\text{mod} \ 31)$
   • $3^{10} = (3^5)^2 = 26^2 = 676 \equiv 26 \ (\text{mod} \ 31)$
   • $3^{15} = 3^{10} \times 3^5 = 26 \times 26 = 676 \equiv 26 \ (\text{mod} \ 31)$
   • $3^{30} = (3^{15})^2 = 26^2 \equiv 1 \ (\text{mod} \ 31)$

So, $k = 30$ is the smallest positive integer for which $3^{30} \equiv 1 \ (\text{mod} \ 31)$.

Thus, the order of the element $3$ in $\mathbb{Z}_{31}^*$ is 30.

Would you like more details or have any questions about this? Here are some additional questions related to the topic:

1. What is the general method for finding the order of an element in a multiplicative group?
2. How does Euler's totient function $\phi(n)$ relate to the group $\mathbb{Z}_n^*$?
3. Can the order of any element in $\mathbb{Z}_p^*$ (where $p$ is prime) ever be greater than $\phi(p)$?
4. What is the significance of the order of an element in group theory?
5. How can we verify that $\mathbb{Z}_{31}^*$ is a cyclic group?

Tip: In a multiplicative group $\mathbb{Z}_p^*$ for a prime $p$, the order of any element always divides $p-1$.