In this exercise we consider the multiplicative group Z ∗ 31. 1. Determine the order of the element 3

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.

---

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.

Learn how to determine the order of the element 3 in the multiplicative group Z*31 with a detailed step-by-step solution. This problem covers group theory and modular arithmetic, exploring the properties of prime numbers. Suitable for advanced high school students and beyond.

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