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The goal of this exercise is to practice finding the inverse modulo m of some (relatively prime) integer n. We will find the inverse of 16 modulo 83, i.e., an integer c such that 16c =1 (mod 83). First we perform the Euclidean algorithm on 16 and 83: 83 = 5* + *5 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(16,83)=D1, i.e., 16 and 83 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 83s + 16t for suitable integers s, t. S = t = when we look at the equation 83s 16t = 1 (mod 83), the multiple of 83 becomes zero and so we get 16t = 1 (mod 83). Hence the multiplicative inverse of 16 modulo 83 is

The goal of this exercise is to practice finding the inverse modulo m of some (relatively prime) integer n. We will find the inverse of 16 modulo 83, i.e., an integer c such that 16c =1 (mod 83). First we perform the Euclidean algorithm on 16 and 83: 83 = 5* + *5 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(16,83)=D1, i.e., 16 and 83 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 83s + 16t for suitable integers s, t. S = t = when we look at the equation 83s 16t = 1 (mod 83), the multiple of 83 becomes zero and so we get 16t = 1 (mod 83). Hence the multiplicative inverse of 16 modulo 83 is

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The goal of this exercise is to practice finding the inverse modulo m of some (relatively prime) integer n. We will find the inverse of 16 modulo 83, i.e., an integer c such that 16c =1 (mod 83). First we perform the Euclidean algorithm on 16 and 83: 83 = 5* + *5 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(16,83)=1, i.e., 16 and 83 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 83s + 16t for suitable integers s, t. S = t = when we look at the equation 83s + 16t = 1 (mod 83), the multiple of 83 becomes zero and so we get 16t = 1 (mod 83). Hence the multiplicative inverse of 16 modulo 83 is
Transcribed Image Text:The goal of this exercise is to practice finding the inverse modulo m of some (relatively prime) integer n. We will find the inverse of 16 modulo 83, i.e., an integer c such that 16c =1 (mod 83). First we perform the Euclidean algorithm on 16 and 83: 83 = 5* + *5 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(16,83)=1, i.e., 16 and 83 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 83s + 16t for suitable integers s, t. S = t = when we look at the equation 83s + 16t = 1 (mod 83), the multiple of 83 becomes zero and so we get 16t = 1 (mod 83). Hence the multiplicative inverse of 16 modulo 83 is
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