What's the multiplicative inverse of 3 in the ring Z base 7 ?

To find the multiplicative inverse of 3 in the ring \( \mathbb{Z}_7 \) (the integers modulo 7), we need to find a number \( x \) such that:

\[
3 \times x \equiv 1 \pmod{7}
\]

This means we want to solve the equation:

\[
3x \equiv 1 \pmod{7}
\]

We can test the integers modulo 7 (0 through 6) to see which satisfies this equation:

- \( 3 \times 1 = 3 \)
- \( 3 \times 2 = 6 \)
- \( 3 \times 3 = 9 \equiv 2 \pmod{7} \)
- \( 3 \times 4 = 12 \equiv 5 \pmod{7} \)
- \( 3 \times 5 = 15 \equiv 1 \pmod{7} \)
- \( 3 \times 6 = 18 \equiv 4 \pmod{7} \)

Thus, \( 3 \times 5 \equiv 1 \pmod{7} \). So, the multiplicative inverse of 3 in \( \mathbb{Z}_7 \) is:

\[
x = 5
\]

### Do you have any questions or would you like further details?

Here are five related questions to consider:
1. What is the multiplicative inverse of 2 in \( \mathbb{Z}_7 \)?
2. How is modular arithmetic used in cryptography?
3. What is the difference between a ring and a field in abstract algebra?
4. Can every element in \( \mathbb{Z}_n \) have a multiplicative inverse?
5. How do you solve linear congruences like \( ax \equiv b \pmod{n} \)?

**Tip:** In any finite ring \( \mathbb{Z}_n \), an element has a multiplicative inverse if and only if it is coprime with \( n \).

What's the multiplicative inverse of 3 in the ring Z base 7?

Learn how to find the multiplicative inverse of 3 in Z7, the ring of integers modulo 7. This detailed guide explains modular arithmetic, key number theory concepts, and provides step-by-step solutions. Suitable for high school students in Grades 10-12.

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