Math Problem Statement

回答4-7

Solution

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

Question 4:

If n,a,n, a, and bb are positive integers for which:

  • nn is odd
  • abmodϕ(n)a \equiv b \mod{\phi(n)}

then 2a2bmodn2^a \equiv 2^b \mod{n}

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

Question 5:

Given S={a,2a,3a,,69a,70a}S = \{a, 2a, 3a, \ldots, 69a, 70a\} with a=3a = 3. Check all that apply.

Let's examine each statement:

  • 3701mod713^{70} \equiv 1 \mod{71} Using Fermat's Little Theorem: ap11modpa^{p-1} \equiv 1 \mod{p} for any integer aa and a prime pp, we get: 3701mod713^{70} \equiv 1 \mod{71} This statement is True.

  • a(2a)(3a)(69a)(70a)70!mod71a \cdot (2a) \cdot (3a) \cdots (69a)(70a) \equiv 70! \mod{71} We need to see if the product a(2a)(70a)a \cdot (2a) \cdots (70a) is congruent to 70!mod7170! \mod{71}. Simplifying this, we note that: (323370)70!mod71(3 \cdot 2 \cdot 3 \cdot \ldots \cdot 3 \cdot 70) \equiv 70! \mod{71} 37070!3^{70} \cdot 70! Since 3701mod713^{70} \equiv 1 \mod{71}, we have: 170!70!mod711 \cdot 70! \equiv 70! \mod{71} This statement is True.

  • SS is a complete residue system mod71\mod{71} A complete residue system modulo 71 would contain every integer from 0 to 70 exactly once. Since SS is 3,6,9,,2103, 6, 9, \ldots, 210, it will cycle through all residues because 33 and 7171 are coprime, thus covering every residue exactly once. This statement is True.

Question 6:

If pp is an odd prime and aa is a positive integer for which a0modpa \equiv 0 \mod{p}, then 2a1modp2^a \equiv 1 \mod{p}.

This statement is True. Since a0modpa \equiv 0 \mod{p}, a=kpa = kp for some integer kk. Therefore: 2a=2kp=(2p)k2^a = 2^{kp} = (2^p)^k By Fermat's Little Theorem, 2p2modp2^p \equiv 2 \mod{p} and 2p11modp2^{p-1} \equiv 1 \mod{p}. Therefore, 2kp1modp2^{kp} \equiv 1 \mod{p}.

Question 7:

If pp and qq are odd primes, then 2pq11modpq2^{pq-1} \equiv 1 \mod{pq}.

This statement is True. It follows from Euler's theorem. For n=pqn = pq, we use ϕ(pq)=(p1)(q1)\phi(pq) = (p-1)(q-1). Euler's theorem tells us: 2ϕ(pq)1modpq2^{\phi(pq)} \equiv 1 \mod{pq} Since ϕ(pq)=(p1)(q1)\phi(pq) = (p-1)(q-1) and pq1pq-1 is one less than pqpq, 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.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic
Euler's Theorem
Fermat's Little Theorem

Formulas

Fermat's Little Theorem: a^(p-1) ≡ 1 (mod p)

Theorems

Euler's Theorem

Suitable Grade Level

Advanced High School