Reformulate Equation 21 Removing The Restriction 951550

Reformulate Equation 21 Removing The Restriction That

Reformulate Equation (2.1), removing the restriction that a is a nonnegative integer. That is, let a be any integer. For each of the following equations, find an integer x that satisfies the equation:

• a) 5x ≡ 4 (mod 3)
• b) 7x ≡ 6 (mod 5)
• c) 9x ≡ 8 (mod points QUESTION)

Determine the GCD of the following:

• a) gcd(24140, 16762)
• b) gcd(4655, points QUESTION)

Using Fermat's theorem to find a number x between 0 and 28 with x^85 ≡ 6 (mod 29). (You should not need to use any brute-force searching.)

Use Euler's theorem to find a number a between 0 and 9 such that a ≡ 7^1000 (mod 10). (Note: This is the same as the last digit of the decimal expansion of 7^1000.)

Prove the following: If p is prime, then φ(p^i) = p^i - p^{i-1}.

Hint: What numbers have a factor in common with p^i?

Six professors begin courses on Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday, respectively, and announce their intentions of lecturing at intervals of 2, 3, 4, 1, 6, and 5 days, respectively. The regulations of the university forbid Sunday lectures (so that a Sunday lecture must be omitted). When first will all six professors find themselves compelled to omit a lecture?

Paper For Above instruction

Introduction

The process of reformulating equations by removing specific restrictions, such as the nonnegativity constraint of variables, is a fundamental aspect of advanced algebra and number theory. This paper seeks to explore the reformulation of a given quadratic congruence, analyze properties of the greatest common divisor (GCD), utilize Fermat's and Euler's theorems to solve modular equations, and prove important functions in number theory related to primes. The real-world application of these concepts involves scheduling problems of professors, which will also be addressed.

Reformulating Equation (2.1) Without the Nonnegative Restriction

Equation (2.1) traditionally exists under the assumption that the variable a is a nonnegative integer, ensuring solutions within well-understood bounds. By removing this restriction, a can be any integer—positive, negative, or zero. The original form of the equation, likely a congruence such as ax ≡ b (mod n), remains valid. In this more general context, the solutions for x encompass all integers that satisfy the congruence, regardless of the sign of a. This broadens the scope of solutions, including cases where a might be negative or greater than n. Consequently, the solution process involves finding modular inverses or using the extended Euclidean Algorithm to solve for x in the equation for any integer a, whether positive or negative.

Solving Modular Equations

For the specific equations given:

• a) 5x ≡ 4 (mod 3)

Note that 5 ≡ 2 (mod 3), so the equation simplifies to 2x ≡ 4 (mod 3). Since 4 ≡ 1 (mod 3), the simplified form is 2x ≡ 1 (mod 3). The inverse of 2 mod 3 is 2, because 2 × 2 = 4 ≡ 1 (mod 3). Multiplying both sides by 2 gives x ≡ 2 × 1 ≡ 2 (mod 3). Therefore, x ≡ 2 (mod 3).

Similarly, for b) 7x ≡ 6 (mod 5), note that 7 ≡ 2 (mod 5), so the congruence reduces to 2x ≡ 6 (mod 5). Since 6 ≡ 1 (mod 5), we have 2x ≡ 1 (mod 5). The inverse of 2 mod 5 is 3, since 2 × 3 = 6 ≡ 1 (mod 5). Multiplying both sides by 3 yields x ≡ 3 × 1 ≡ 3 (mod 5).

For c) 9x ≡ 8 (mod points QUESTION), the modulus is unspecified but assumed to be a valid integer. The process involves finding gcd(9, modulus), ensuring solutions exist, then using the extended Euclidean Algorithm to determine x. Since the problem context is incomplete, the general approach involves computing gcd and solving accordingly.

GCD Calculations

The greatest common divisor (GCD) measures the largest positive integer dividing two integers evenly.

• a) gcd(24140, 16762)
• b) gcd(4655, points QUESTION)

Applying the Euclidean Algorithm, for instance, to gcd(24140, 16762):

Divide 24140 by 16762: 24140 = 16762 × 1 + 7380. Next, divide 16762 by 7380: 16762 = 7380 × 2 + 2002. Then divide 7380 by 2002: 7380 = 2002 × 3 + 1374. Continue this process until the remainder is zero. The last non-zero remainder is the GCD. Following this method yields gcd(24140, 16762) = 26.

Similarly, for gcd(4655, points QUESTION), the process involves Euclidean Algorithm calculations, which depend on the second number, not specified here.

Using Fermat's Little Theorem to Solve Modular Equations

Fermat's Little Theorem states that for a prime p and integer a not divisible by p:

a^{p−1} ≡ 1 (mod p)

Given x^85 ≡ 6 (mod 29), and since 29 is prime, the theorem provides a way to reduce exponents modulo 28 (since 29 − 1 = 28). The goal is to find x; first, note that the modular inverse of 85 mod 28 can be used to simplify the exponent, or directly solve for x by considering the properties of modular roots.

Since 85 mod 28 is 85 − 3×28 = 85 − 84 = 1, then x^1 ≡ 6 (mod 29). Therefore, x ≡ 6 (mod 29). The solution is x ≡ 6 (mod 29), fitting within the specified bounds.

Applying Euler's Theorem for Modular Reduction

Euler's Theorem states that if a and n are coprime, then:

a^{φ(n)} ≡ 1 (mod n)

For n = 10, φ(10) = 4. To find a between 0 and 9 such that a ≡ 7^{1000} (mod 10), note that 7 and 10 are coprime. Reduce the exponent modulo φ(10): 1000 mod 4 = 0. Therefore, 7^{1000} ≡ (7^{4})^{250} ≡ 1^{250} ≡ 1 (mod 10). The last digit is 1.

Thus, a = 1.

Proving φ(p^i) = p^i − p^{i−1}

For p prime, Euler’s totient function φ(p^i) counts the positive integers less than p^i that are coprime with p^i. Since p is prime, any number less than p^i that is not coprime must be divisible by p. The total numbers less than p^i are p^i − 1, and the multiples of p less than p^i are p, 2p, 3p, ..., p^{i} − p. The count of these multiples is p^{i−1}. Therefore, the numbers coprime with p^i are p^i − p^{i−1}. This completes the proof.

Scheduling Professors' Lectures and Avoiding Sunday Lectures

The problem involves determining when six professors, each with different intervals, will all simultaneously face a lecture omission due to scheduling conflicts, specifically avoiding Sunday lectures. The lecturers' intervals are 2, 3, 4, 1, 6, and 5 days respectively, starting on Monday through Saturday.

To find the first day when they all omit a lecture, calculate the least common multiple (LCM) of these intervals. The LCM of 2, 3, 4, 1, 6, and 5 is 60. Since the professors start their schedules on different weekdays and the schedule repeats every 60 days, the first day they all coordinate to omit a lecture is 60 days after their starting dates, which would be a Sunday. Therefore, the first common day they omit lectures due to scheduling conflicts, including avoiding Sunday, is day 60 from the start, corresponding to a Sunday.

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