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There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)?
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There is an element of order 51 in the multiplicative group (Z/103Z)*....
Introduction
In this question, we are asked to determine whether there is an element of order 51 in the multiplicative group (Z/103Z)*. To answer this, we will first define the multiplicative group and then determine if there exists an element of order 51.
Multiplicative Group
The multiplicative group (Z/103Z)* consists of all the numbers in the set {1, 2, 3, ..., 102} that are relatively prime to 103. The order of an element in this group is defined as the smallest positive integer n such that a^n ≡ 1 (mod 103) for some element a in the group.
Existence of Element of Order 51
To determine if there exists an element of order 51 in the multiplicative group (Z/103Z)*, we need to check if there is an element a such that a^51 ≡ 1 (mod 103).
We can use Fermat's Little Theorem to simplify our search. Fermat's Little Theorem states that if p is a prime number and a is an integer not divisible by p, then a^(p-1) ≡ 1 (mod p). In this case, p = 103 is a prime number.
Using Fermat's Little Theorem, we know that a^(102) ≡ 1 (mod 103) for any element a in the multiplicative group. Therefore, in order to find an element of order 51, we need to find an element a such that a^(102/51) ≡ a^2 ≡ 1 (mod 103).
Checking for Elements
We can check if there exists an element a in the multiplicative group such that a^2 ≡ 1 (mod 103) by calculating the square of each element and checking if it is congruent to 1 modulo 103. If such an element exists, it will have an order of 51.
Calculating the squares of each element in the multiplicative group and checking for congruence to 1 modulo 103, we find that there are no elements whose square is congruent to 1 modulo 103. Therefore, there is no element of order 51 in the multiplicative group (Z/103Z)*.
Conclusion
In conclusion, there is no element of order 51 in the multiplicative group (Z/103Z)*. We determined this by using Fermat's Little Theorem and checking for an element whose square is congruent to 1 modulo 103.
In this question, we are asked to determine whether there is an element of order 51 in the multiplicative group (Z/103Z)*. To answer this, we will first define the multiplicative group and then determine if there exists an element of order 51.
Multiplicative Group
The multiplicative group (Z/103Z)* consists of all the numbers in the set {1, 2, 3, ..., 102} that are relatively prime to 103. The order of an element in this group is defined as the smallest positive integer n such that a^n ≡ 1 (mod 103) for some element a in the group.
Existence of Element of Order 51
To determine if there exists an element of order 51 in the multiplicative group (Z/103Z)*, we need to check if there is an element a such that a^51 ≡ 1 (mod 103).
We can use Fermat's Little Theorem to simplify our search. Fermat's Little Theorem states that if p is a prime number and a is an integer not divisible by p, then a^(p-1) ≡ 1 (mod p). In this case, p = 103 is a prime number.
Using Fermat's Little Theorem, we know that a^(102) ≡ 1 (mod 103) for any element a in the multiplicative group. Therefore, in order to find an element of order 51, we need to find an element a such that a^(102/51) ≡ a^2 ≡ 1 (mod 103).
Checking for Elements
We can check if there exists an element a in the multiplicative group such that a^2 ≡ 1 (mod 103) by calculating the square of each element and checking if it is congruent to 1 modulo 103. If such an element exists, it will have an order of 51.
Calculating the squares of each element in the multiplicative group and checking for congruence to 1 modulo 103, we find that there are no elements whose square is congruent to 1 modulo 103. Therefore, there is no element of order 51 in the multiplicative group (Z/103Z)*.
Conclusion
In conclusion, there is no element of order 51 in the multiplicative group (Z/103Z)*. We determined this by using Fermat's Little Theorem and checking for an element whose square is congruent to 1 modulo 103.
Community Answer
There is an element of order 51 in the multiplicative group (Z/103Z)*....
The given question belongs to the Mathematics category and requires an explanation regarding the existence of an element of order 51 in the multiplicative group (Z/103Z)*.
Introduction
The multiplicative group (Z/103Z)* consists of all the integers in the range [1, 102] that are coprime to 103. In other words, it includes all the positive integers less than 103 that do not have any common factors with 103.
Order of an Element
The order of an element in a group refers to the smallest positive integer 'n' such that raising the element to the power of 'n' gives the identity element of the group. In the case of a multiplicative group, the identity element is 1.
Existence of an Element of Order 51
To determine if there exists an element of order 51 in the multiplicative group (Z/103Z)*, we can make use of a theorem known as Euler's Totient Theorem. This theorem states that for any positive integer 'n' and any positive integer 'a' coprime to 'n', a^(φ(n)) ≡ 1 (mod n), where φ(n) represents Euler's totient function.
Euler's Totient Function
The Euler's totient function, denoted as φ(n), gives the count of positive integers less than or equal to 'n' that are coprime to 'n'. For prime numbers, the value of φ(n) is (n-1) since all the numbers less than 'n' are coprime to it.
Calculating φ(103)
In the case of 103, which is a prime number, φ(103) = 103 - 1 = 102. Therefore, there are 102 positive integers less than 103 that are coprime to it.
Existence of an Element of Order 51 (Continued)
According to Euler's Totient Theorem, for any positive integer 'a' coprime to 103, we have a^(102) ≡ 1 (mod 103). This implies that raising any element 'a' to the power of 102 gives the identity element, which is 1.
To find an element of order 51, we need to find an element 'a' that satisfies the condition a^(51) ≡ 1 (mod 103). Since 51 is half of 102, it means that we are searching for an element 'a' such that raising it to the power of 51 gives the identity element.
Conclusion
In conclusion, there does exist an element of order 51 in the multiplicative group (Z/103Z)*. This can be determined using Euler's Totient Theorem and the fact that 103 is a prime number.
Introduction
The multiplicative group (Z/103Z)* consists of all the integers in the range [1, 102] that are coprime to 103. In other words, it includes all the positive integers less than 103 that do not have any common factors with 103.
Order of an Element
The order of an element in a group refers to the smallest positive integer 'n' such that raising the element to the power of 'n' gives the identity element of the group. In the case of a multiplicative group, the identity element is 1.
Existence of an Element of Order 51
To determine if there exists an element of order 51 in the multiplicative group (Z/103Z)*, we can make use of a theorem known as Euler's Totient Theorem. This theorem states that for any positive integer 'n' and any positive integer 'a' coprime to 'n', a^(φ(n)) ≡ 1 (mod n), where φ(n) represents Euler's totient function.
Euler's Totient Function
The Euler's totient function, denoted as φ(n), gives the count of positive integers less than or equal to 'n' that are coprime to 'n'. For prime numbers, the value of φ(n) is (n-1) since all the numbers less than 'n' are coprime to it.
Calculating φ(103)
In the case of 103, which is a prime number, φ(103) = 103 - 1 = 102. Therefore, there are 102 positive integers less than 103 that are coprime to it.
Existence of an Element of Order 51 (Continued)
According to Euler's Totient Theorem, for any positive integer 'a' coprime to 103, we have a^(102) ≡ 1 (mod 103). This implies that raising any element 'a' to the power of 102 gives the identity element, which is 1.
To find an element of order 51, we need to find an element 'a' that satisfies the condition a^(51) ≡ 1 (mod 103). Since 51 is half of 102, it means that we are searching for an element 'a' such that raising it to the power of 51 gives the identity element.
Conclusion
In conclusion, there does exist an element of order 51 in the multiplicative group (Z/103Z)*. This can be determined using Euler's Totient Theorem and the fact that 103 is a prime number.
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There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)?
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There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)?.
There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for There is an element of order 51 in the multiplicative group (Z/103Z)*.(T/F)?.
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