Prove that the order of 3 modulo 19 is a divisor of 18.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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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.
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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