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

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

In this case, the element is $g = 3$, and the group $\mathbb{Z}_{31}^*$ consists of all the integers from 1 to 30 under multiplication modulo 31.

Step 1: Determine the size of $\mathbb{Z}_{31}^*$

Since 31 is a prime number, the group $\mathbb{Z}_{31}^*$ has $31 - 1 = 30$ elements. Therefore, the order of the group $\mathbb{Z}_{31}^*$ is 30.

Step 2: Find the order of the element 3

The order of the element 3 is the smallest positive integer $k$ such that $3^k \equiv 1 \ (\text{mod}\ 31)$.

To find $k$, we can try successive powers of 3 modulo 31 until we get 1.

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

$3^2 = 9 \ (\text{mod}\ 31)$

$3^3 = 27 \ (\text{mod}\ 31)$

$3^4 = 3 \times 27 = 81 \equiv 19 \ (\text{mod}\ 31)$

$3^5 = 3 \times 19 = 57 \equiv 26 \ (\text{mod}\ 31)$

$3^6 = 3 \times 26 = 78 \equiv 16 \ (\text{mod}\ 31)$

$3^7 = 3 \times 16 = 48 \equiv 17 \ (\text{mod}\ 31)$

$3^8 = 3 \times 17 = 51 \equiv 20 \ (\text{mod}\ 31)$

$3^9 = 3 \times 20 = 60 \equiv 29 \ (\text{mod}\ 31)$

$3^{10} = 3 \times 29 = 87 \equiv 25 \ (\text{mod}\ 31)$

$3^{11} = 3 \times 25 = 75 \equiv 13 \ (\text{mod}\ 31)$

$3^{12} = 3 \times 13 = 39 \equiv 8 \ (\text{mod}\ 31)$

$3^{13} = 3 \times 8 = 24 \ (\text{mod}\ 31)$

$3^{14} = 3 \times 24 = 72 \equiv 10 \ (\text{mod}\ 31)$

$3^{15} = 3 \times 10 = 30 \ (\text{mod}\ 31)$

$3^{16} = 3 \times 30 = 90 \equiv 28 \ (\text{mod}\ 31)$

$3^{17} = 3 \times 28 = 84 \equiv 22 \ (\text{mod}\ 31)$

$3^{18} = 3 \times 22 = 66 \equiv 4 \ (\text{mod}\ 31)$

$3^{19} = 3 \times 4 = 12 \ (\text{mod}\ 31)$

$3^{20} = 3 \times 12 = 36 \equiv 5 \ (\text{mod}\ 31)$

$3^{21} = 3 \times 5 = 15 \ (\text{mod}\ 31)$

$3^{22} = 3 \times 15 = 45 \equiv 14 \ (\text{mod}\ 31)$

$3^{23} = 3 \times 14 = 42 \equiv 11 \ (\text{mod}\ 31)$

$3^{24} = 3 \times 11 = 33 \equiv 2 \ (\text{mod}\ 31)$

$3^{25} = 3 \times 2 = 6 \ (\text{mod}\ 31)$

$3^{26} = 3 \times 6 = 18 \ (\text{mod}\ 31)$

$3^{27} = 3 \times 18 = 54 \equiv 23 \ (\text{mod}\ 31)$

$3^{28} = 3 \times 23 = 69 \equiv 7 \ (\text{mod}\ 31)$

$3^{29} = 3 \times 7 = 21 \ (\text{mod}\ 31)$

$3^{30} = 3 \times 21 = 63 \equiv 1 \ (\text{mod}\ 31)$

The first occurrence of 1 is when $k = 30$.

Conclusion:

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

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Do you have any questions or would you like further details? Here are some related questions you might find interesting:

1. How do you determine the order of other elements in $\mathbb{Z}_{31}^*$?
2. What is the significance of the order of an element in a group?
3. How does the order of an element relate to the structure of the group?
4. Can you find a generator for $\mathbb{Z}_{31}^*$?
5. What is the connection between the order of an element and its cyclic subgroup?

Tip: When working with groups like $\mathbb{Z}_{p}^*$, where $p$ is a prime number, the order of an element always divides the order of the group (which is $p-1$). This property can help in narrowing down the possible values for the order of any given element.