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Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ?
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Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo...
Solution:
To find the value of x in the given group G, we need to solve the equation x^2 ≡ 26 (mod 55) for x > 30.
1. Prime factorization of 55:
To understand the properties of the given group, let's start by finding the prime factorization of 55.
55 = 5 * 11
2. Euler's Totient function (φ):
Since we are dealing with a group under multiplication modulo 55, we need to calculate the Euler's Totient function (φ) of 55, denoted as φ(55). Euler's Totient function gives the count of numbers less than 55 that are coprime with 55.
3. Calculation of φ(55):
Since 55 = 5 * 11, and both 5 and 11 are prime factors, we can use the formula for calculating φ(n) where n is a prime power.
φ(5) = 5 - 1 = 4
φ(11) = 11 - 1 = 10
Since 5 and 11 are coprime, we can use the property φ(m * n) = φ(m) * φ(n) for coprime numbers.
φ(55) = φ(5) * φ(11) = 4 * 10 = 40
4. Properties of the given group G:
Since G is a group under multiplication modulo 55, it has the following properties:
- It is closed under multiplication modulo 55.
- Every element in G has an inverse in G.
- The order of the group is given by φ(55) = 40.
5. Solving the equation x^2 ≡ 26 (mod 55):
To find the value of x, we need to solve the given equation x^2 ≡ 26 (mod 55) in the group G.
Since the order of G is 40, the equation can be written as (x^2)^20 ≡ 26^20 (mod 55).
Using Fermat's Little Theorem, which states that if p is a prime number and a is any integer not divisible by p, then a^(p-1) ≡ 1 (mod p), we can simplify the equation as follows:
(x^2)^20 ≡ 26^20 (mod 55)
x^40 ≡ 1 (mod 55)
Now, we need to find the value of x in G that satisfies the equation x^40 ≡ 1 (mod 55).
6. Finding x:
Since x > 30, we can narrow down our search by considering the values of x from 31 to 55.
By checking each value of x from 31 to 55, we can find the value of x that satisfies the equation x^40 ≡ 1 (mod 55).
After checking the values, we find that x = 36 satisfies the equation x^40 ≡ 1 (mod 55).
Therefore, x = 36 is the value of x in the given group G that satisfies the equation x^2 ≡ 26 (mod 55) and x > 30.
To find the value of x in the given group G, we need to solve the equation x^2 ≡ 26 (mod 55) for x > 30.
1. Prime factorization of 55:
To understand the properties of the given group, let's start by finding the prime factorization of 55.
55 = 5 * 11
2. Euler's Totient function (φ):
Since we are dealing with a group under multiplication modulo 55, we need to calculate the Euler's Totient function (φ) of 55, denoted as φ(55). Euler's Totient function gives the count of numbers less than 55 that are coprime with 55.
3. Calculation of φ(55):
Since 55 = 5 * 11, and both 5 and 11 are prime factors, we can use the formula for calculating φ(n) where n is a prime power.
φ(5) = 5 - 1 = 4
φ(11) = 11 - 1 = 10
Since 5 and 11 are coprime, we can use the property φ(m * n) = φ(m) * φ(n) for coprime numbers.
φ(55) = φ(5) * φ(11) = 4 * 10 = 40
4. Properties of the given group G:
Since G is a group under multiplication modulo 55, it has the following properties:
- It is closed under multiplication modulo 55.
- Every element in G has an inverse in G.
- The order of the group is given by φ(55) = 40.
5. Solving the equation x^2 ≡ 26 (mod 55):
To find the value of x, we need to solve the given equation x^2 ≡ 26 (mod 55) in the group G.
Since the order of G is 40, the equation can be written as (x^2)^20 ≡ 26^20 (mod 55).
Using Fermat's Little Theorem, which states that if p is a prime number and a is any integer not divisible by p, then a^(p-1) ≡ 1 (mod p), we can simplify the equation as follows:
(x^2)^20 ≡ 26^20 (mod 55)
x^40 ≡ 1 (mod 55)
Now, we need to find the value of x in G that satisfies the equation x^40 ≡ 1 (mod 55).
6. Finding x:
Since x > 30, we can narrow down our search by considering the values of x from 31 to 55.
By checking each value of x from 31 to 55, we can find the value of x that satisfies the equation x^40 ≡ 1 (mod 55).
After checking the values, we find that x = 36 satisfies the equation x^40 ≡ 1 (mod 55).
Therefore, x = 36 is the value of x in the given group G that satisfies the equation x^2 ≡ 26 (mod 55) and x > 30.
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Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ?
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Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ?.
Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let G ={n£N:n≤55,gcd(n,55)=1} be the group under multiplication modulo 55. Let x£G be such that x^2=26 and x>30.then x is equal to ?.
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