To calculate the order of the element $3$ in the multiplicative group of integers modulo $31$, we need to find the smallest positive integer $k$ such that:

$3^k \equiv 1 \pmod{31}$

Steps to find the order:

1. List possible orders: The order $k$ must divide the order of the group. The group of integers modulo 31, denoted $(\mathbb{Z}/31\mathbb{Z})^\times$, has order $31 - 1 = 30$.

   Therefore, the possible values for $k$ are the divisors of 30: $1, 2, 3, 5, 6, 10, 15, 30$.

2. Calculate powers of 3 modulo 31:

   • $3^1 \equiv 3 \pmod{31}$
   • $3^2 \equiv 9 \pmod{31}$
   • $3^3 \equiv 27 \pmod{31}$
   • $3^5 \equiv 243 \equiv 26 \pmod{31}$
   • $3^6 \equiv 26 \times 3 \equiv 78 \equiv 16 \pmod{31}$
   • $3^{10} \equiv 16 \times 16 \equiv 256 \equiv 8 \pmod{31}$
   • $3^{15} \equiv 8 \times 26 \equiv 208 \equiv 22 \pmod{31}$
   • $3^{30} \equiv (3^{15})^2 \equiv 22^2 \equiv 484 \equiv 1 \pmod{31}$

   As seen from the calculations above, $3^{30} \equiv 1 \pmod{31}$, and none of the smaller powers of 3 equal 1 modulo 31.

Conclusion:

The smallest $k$ for which $3^k \equiv 1 \pmod{31}$ is $k = 30$. Therefore, the order of the element 3 in the multiplicative group modulo 31 is 30.

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Would you like more details on this topic or have any questions?

Here are 5 related questions you might be interested in:

1. What is the multiplicative group modulo $p$ where $p$ is a prime number?
2. How do you find the order of an element in any group?
3. What are primitive roots, and how do they relate to this problem?
4. How would the approach differ if $31$ were not a prime number?
5. Can every element in a multiplicative group modulo $p$ be a generator?

Tip: The order of an element in a group is always a divisor of the group's order. This is a direct consequence of Lagrange's theorem in group theory.