https://file.aiquickdraw.com/mathbot/file/1720976113013qbad7g5p.png

Let's solve questions 4 to 7 step by step.

### Question 4:
If \( n, a, \) and \( b \) are positive integers for which:
- \( n \) is odd
- \( a \equiv b \mod{\phi(n)} \)

then \( 2^a \equiv 2^b \mod{n} \)

This statement is True. It is a direct application of Euler's theorem which states that if \( a \equiv b \mod{\phi(n)} \), then \( x^a \equiv x^b \mod{n} \) for any integer \( x \) that is coprime with \( n \).

### Question 5:
Given \( S = \{a, 2a, 3a, \ldots, 69a, 70a\} \) with \( a = 3 \).
Check all that apply.

Let's examine each statement:

- \( 3^{70} \equiv 1 \mod{71} \)
  Using Fermat's Little Theorem: \( a^{p-1} \equiv 1 \mod{p} \) for any integer \( a \) and a prime \( p \), we get:
  \[ 3^{70} \equiv 1 \mod{71} \]
  This statement is True.

- \( a \cdot (2a) \cdot (3a) \cdots (69a)(70a) \equiv 70! \mod{71} \)
  We need to see if the product \( a \cdot (2a) \cdots (70a) \) is congruent to \( 70! \mod{71} \). Simplifying this, we note that:
  \[ (3 \cdot 2 \cdot 3 \cdot \ldots \cdot 3 \cdot 70) \equiv 70! \mod{71} \]
  \[ 3^{70} \cdot 70! \]
  Since \( 3^{70} \equiv 1 \mod{71} \), we have:
  \[ 1 \cdot 70! \equiv 70! \mod{71} \]
  This statement is True.

- \( S \) is a complete residue system \( \mod{71} \)
  A complete residue system modulo 71 would contain every integer from 0 to 70 exactly once. Since \( S \) is \( 3, 6, 9, \ldots, 210 \), it will cycle through all residues because \( 3 \) and \( 71 \) are coprime, thus covering every residue exactly once.
  This statement is True.

### Question 6:
If \( p \) is an odd prime and \( a \) is a positive integer for which \( a \equiv 0 \mod{p} \), then \( 2^a \equiv 1 \mod{p} \).

This statement is True. Since \( a \equiv 0 \mod{p} \), \( a = kp \) for some integer \( k \). Therefore:
\[ 2^a = 2^{kp} = (2^p)^k \]
By Fermat's Little Theorem, \( 2^p \equiv 2 \mod{p} \) and \( 2^{p-1} \equiv 1 \mod{p} \). Therefore, \( 2^{kp} \equiv 1 \mod{p} \).

### Question 7:
If \( p \) and \( q \) are odd primes, then \( 2^{pq-1} \equiv 1 \mod{pq} \).

This statement is True. It follows from Euler's theorem. For \( n = pq \), we use \( \phi(pq) = (p-1)(q-1) \). Euler's theorem tells us:
\[ 2^{\phi(pq)} \equiv 1 \mod{pq} \]
Since \( \phi(pq) = (p-1)(q-1) \) and \( pq-1 \) is one less than \( pq \), it satisfies the theorem.

Would you like more details on any of these questions or have any further questions? Here are five more questions you might be interested in:

1. How can Euler's theorem be applied to cryptography?
2. What is Fermat's Little Theorem and how is it used?
3. How can we determine if a set forms a complete residue system?
4. What are the properties of modular arithmetic?
5. Can you provide examples of calculating modular exponentiation?

### Tip:
Understanding the principles of modular arithmetic and number theory can greatly simplify solving problems involving congruences and the Euler's totient function.

Explore the application of Euler's Theorem and Fermat's Little Theorem in number theory problems. This detailed explanation covers questions involving modular arithmetic, congruences, and prime numbers. Suitable for advanced high school students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation