1.30. For each of the following primes p and numbers a, compute a-¹ mod p in two ways: (i) Use the extended Euclidean algorithm. (ii) Use the fast power algorithm and Fermat's little theorem. (See Example 1.28.) (a) p = 47 and a = 11. (b) p= 587 and a = 345. (c) p = 104801 and a = 78467.
1.30. For each of the following primes p and numbers a, compute a-¹ mod p in two ways: (i) Use the extended Euclidean algorithm. (ii) Use the fast power algorithm and Fermat's little theorem. (See Example 1.28.) (a) p = 47 and a = 11. (b) p= 587 and a = 345. (c) p = 104801 and a = 78467.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:### Exercise 1.30
For each of the following primes \( p \) and numbers \( a \), compute \( a^{-1} \mod p \) in two ways:
1. **Using the extended Euclidean algorithm.**
2. **Using the fast power algorithm and Fermat’s little theorem.** (See Example 1.28.)
#### Given:
(a) \( p = 47 \) and \( a = 11 \).
(b) \( p = 587 \) and \( a = 345 \).
(c) \( p = 104801 \) and \( a = 78467 \).
![### Example 1.28. Inverse Computation Modulo
In this example, we compute the inverse of 7814 modulo 17449 in two different methods.
#### First Method
Initially, we calculate the inverse using exponentiation:
\[ 7814^{-1} \equiv 7814^{17447} \equiv 1284 \pmod{17449}. \]
#### Second Method
Next, we use the extended Euclidean algorithm to solve:
\[ 7814u + 17449v = 1. \]
The solution is \( (u, v) = (1284, -575) \), thus:
\[ 7814^{-1} \equiv 1284 \pmod{17449}. \]
This demonstrates two approaches to finding the modular inverse of a number.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c5ad030-3ec8-4fd2-8d64-821b0d0d0877%2F8315d7e7-523a-4db0-bf40-b2df53e3e752%2Ftia5zd8_processed.png&w=3840&q=75)
Transcribed Image Text:### Example 1.28. Inverse Computation Modulo
In this example, we compute the inverse of 7814 modulo 17449 in two different methods.
#### First Method
Initially, we calculate the inverse using exponentiation:
\[ 7814^{-1} \equiv 7814^{17447} \equiv 1284 \pmod{17449}. \]
#### Second Method
Next, we use the extended Euclidean algorithm to solve:
\[ 7814u + 17449v = 1. \]
The solution is \( (u, v) = (1284, -575) \), thus:
\[ 7814^{-1} \equiv 1284 \pmod{17449}. \]
This demonstrates two approaches to finding the modular inverse of a number.
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