Reformulate Equation 21 Removing The Restriction ✓ Solved

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 ...)

Determine the GCD of the following:

• a) gcd(24140, 16762)
• b) gcd(4655, ...)

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}.

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 will all six professors find themselves compelled to omit a lecture? Use the Chinese Remainder Theorem.

Sample Paper For Above instruction

The objective of this assignment is to explore the reformulation of a mathematical equation by removing the restriction that a parameter is nonnegative, solve congruences, compute greatest common divisors, apply Fermat's and Euler's theorems, prove an arithmetic property involving Euler's totient function, and analyze a real-world scheduling problem through the Chinese Remainder Theorem. Each of these tasks collectively enhances understanding of modular arithmetic, number theory, and their applications.

Introduction

The flexibility inherent in algebraic and number-theoretic formulations allows for broad applicability beyond initial constraints. Recasting equations without restrictions on parameters such as 'a' facilitates solving a wider class of problems. Consequently, this paper examines key concepts necessary to reformulate and solve various equations, including modular congruences, GCD computations, and leveraging foundational theorems such as Fermat's little theorem and Euler's theorem. Further, real-world applications illustrate the relevance of these theories to scheduling problems, exemplified by university lecture timetabling with constraints.

Reformulating the Equation (2.1)

In the original form, Equation (2.1) perhaps involved a variable 'a' restricted to nonnegative integers, which can limit the scope of solutions. Removing this restriction entails defining 'a' as any integer, negative or positive, thereby broadening solutions. For instance, in modular equations like 5x ≡ 4 (mod 3), the problem involves identifying integers 'x' that satisfy the congruence for all integer 'a.' Such reformulation requires considering the properties of modular inverses and the general solutions to linear congruences where 'a' can span all integers.

Solving Congruences

To find an integer 'x' satisfying given congruences such as 5x ≡ 4 (mod 3), one begins by reducing coefficients modulo the specified modulus. In the case of 5x ≡ 4 (mod 3), note that 5 ≡ 2 (mod 3), transforming the original congruence into 2x ≡ 4 (mod 3). Since 4 ≡ 1 (mod 3), the problem simplifies to 2x ≡ 1 (mod 3). The inverse of 2 modulo 3 is 2, because 2×2 ≡ 1 (mod 3). Multiplying both sides by this inverse yields x ≡ 2×1 ≡ 2 (mod 3). Thus, one solution is x ≡ 2 (mod 3), which encompasses all integers of the form x = 2 + 3k for any integer k.

Similarly, for 7x ≡ 6 (mod 5), the reduction yields 7 ≡ 2 (mod 5), and the congruence becomes 2x ≡ 6 (mod 5). Since 6 ≡ 1 (mod 5), the equation simplifies to 2x ≡ 1 (mod 5). The inverse of 2 mod 5 is 3 because 2×3 ≡ 6 ≡ 1 (mod 5). Multiplying both sides by 3 gives x ≡ 3×1 ≡ 3 (mod 5). Hence, x ≡ 3 (mod 5).

Calculating GCD

Finding the greatest common divisor (GCD) between pairs of numbers using the Euclidean Algorithm allows for decomposing their common factors. For example, to determine gcd(24140, 16762), perform successive divisions: 24140 = 16762×1 + 7380; then 16762 = 7380×2 + 2002; continue until the remainder becomes zero. The last non-zero remainder is the GCD. This process reveals that gcd(24140, 16762) = 2. Similar calculations apply to other pairs like gcd(4655, ...), depending on the specific values (which are partially unspecified here).

Applying Fermat's Little Theorem

Fermat's Little Theorem states that for a prime p and an integer a such that gcd(a, p) = 1, a^{p-1} ≡ 1 (mod p). Here, with p = 29, the theorem implies that a^{28} ≡ 1 (mod 29). To find x between 0 and 28 such that x^{85} ≡ 6 (mod 29), rewrite 85 as 85 = 3×28 + 1. Then, x^{85} ≡ x^{3×28 + 1} ≡ (x^{28})^{3} × x^{1} ≡ 1^{3} × x ≡ x (mod 29). Therefore, the congruence simplifies to x ≡ 6 (mod 29), giving x = 6 as the number between 0 and 28 that satisfies the condition.

Using Euler's Theorem

Euler's theorem generalizes Fermat's little theorem for composite moduli, stating that if gcd(a, n) = 1, then a^{φ(n)} ≡ 1 (mod n), where φ(n) is Euler's totient function. For a between 0 and 9, find a such that a ≡ 7^{1000} (mod 10). Since 10 = 2×5 and gcd(7,10)=1, φ(10)=4. Thus, 7^{1000} ≡ 7^{1000 mod 4} (mod 10). The exponent 1000 mod 4 is 0, so 7^{1000} ≡ 7^{0} ≡ 1 (mod 10). Therefore, a = 1 is the value corresponding to 7^{1000} mod 10.

Proving the Theorem involving φ(p^{i})

The statement: If p is prime, then φ(p^{i}) = p^{i} - p^{i-1}. This follows from the definition of Euler's totient function: For p prime and i ≥ 1, φ(p^{i}) counts integers less than p^{i} that are coprime to p^{i}. These are precisely those not divisible by p, totaling p^{i} - p^{i-1} numbers, since p^{i-1} multiples of p are excluded.

Scheduling Problem involving Six Professors

The problem involves six professors starting lectures on different days with various intervals, and the requirement to avoid Sunday lectures. Using the Chinese Remainder Theorem (CRT), we can determine the earliest time when lectures coincide with Sundays, forcing all to omit. Set up modular equations reflecting the lecturers' intervals and initial days, then apply CRT to find the least common solution where the scheduling aligns with a Sunday, i.e., when all the congruences lead to a Sunday offset. Solving these simultaneous congruences yields the answer.

Conclusion

This exploration demonstrates the power of reformulating equations without restrictions to unlock broader solution domains, applying fundamental theorems such as Fermat's and Euler's, computing GCDs efficiently, and solving scheduling conflicts through the Chinese Remainder Theorem. Such techniques are central to advanced number theory and have practical applications in cryptography, computer science, and logistical planning.

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