Prove that the order of 3 modulo 19 is a divisor of 18.
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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![**Problem Statement:**
Prove that the order of 3 modulo 19 is a divisor of 18.
**Explanation:**
To prove this, we need to understand the concept of the order of an integer modulo \( n \). The order of an integer \( a \) modulo \( n \) is the smallest positive integer \( k \) such that:
\[ a^k \equiv 1 \pmod{n} \]
For the given problem, we are interested in the order of 3 modulo 19. According to Fermat's Little Theorem, since 19 is a prime, for any integer \( a \) that is not divisible by 19:
\[ a^{18} \equiv 1 \pmod{19} \]
This means that the order of 3 modulo 19 must be a divisor of 18, since 18 is the smallest integer for which 3 raised to that power is congruent to 1 modulo 19. The divisors of 18 are \( 1, 2, 3, 6, 9, \) and \( 18 \).
To find the exact order, you would compute powers of 3 modulo 19 until you reach 1, ensuring to check the smallest power that satisfies this condition.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30761ad5-6d22-4ff4-adb6-4f166a7ab52a%2F2d6f9ba3-c0ba-493e-8eff-aabf6faefef7%2F8z18tk4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Prove that the order of 3 modulo 19 is a divisor of 18.
**Explanation:**
To prove this, we need to understand the concept of the order of an integer modulo \( n \). The order of an integer \( a \) modulo \( n \) is the smallest positive integer \( k \) such that:
\[ a^k \equiv 1 \pmod{n} \]
For the given problem, we are interested in the order of 3 modulo 19. According to Fermat's Little Theorem, since 19 is a prime, for any integer \( a \) that is not divisible by 19:
\[ a^{18} \equiv 1 \pmod{19} \]
This means that the order of 3 modulo 19 must be a divisor of 18, since 18 is the smallest integer for which 3 raised to that power is congruent to 1 modulo 19. The divisors of 18 are \( 1, 2, 3, 6, 9, \) and \( 18 \).
To find the exact order, you would compute powers of 3 modulo 19 until you reach 1, ensuring to check the smallest power that satisfies this condition.
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