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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 i.e., an integer c such that 11c = 1 (mod 82). First we perform the Euclidean algorithm on 11 and 82: 82 = 7* + *2 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(11,82)=1, i.e., 11 and 82 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 = 82s + 11t for suitable integers s, t. S = t = when we look at the equation 82s + 11t = 1 (mod 82), the multiple of 82 becomes zero and so we get (mod 82) ultiplicatiu 80 is

The goal of this exercise is to practice finding the inverse modulo m of some (relatively prime) integer n. We will find the i.e., an integer c such that 11c = 1 (mod 82). First we perform the Euclidean algorithm on 11 and 82: 82 = 7* + *2 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(11,82)=1, i.e., 11 and 82 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 = 82s + 11t for suitable integers s, t. S = t = when we look at the equation 82s + 11t = 1 (mod 82), the multiple of 82 becomes zero and so we get (mod 82) ultiplicatiu 80 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 11 modulo 82, i.e., an integer c such that 11c = 1 (mod 82). First we perform the Euclidean algorithm on 11 and 82: 82 = 7* + = *2 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(11,82)=1, i.e., 11 and 82 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 = S = 82s + 11t for suitable integers s, t. t = when we look at the equation 82s + 11t = 1 (mod 82), the multiple of 82 becomes zero and so we get 11t 1 (mod 82). Hence the multiplicative inverse of 11 modulo 82 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 11 modulo 82, i.e., an integer c such that 11c = 1 (mod 82). First we perform the Euclidean algorithm on 11 and 82: 82 = 7* + = *2 +1 [Note your answers on the second row should match the ones on the first row.] Thus gcd(11,82)=1, i.e., 11 and 82 are relatively prime. Now we run the Euclidean algorithm backwards to write 1 = S = 82s + 11t for suitable integers s, t. t = when we look at the equation 82s + 11t = 1 (mod 82), the multiple of 82 becomes zero and so we get 11t 1 (mod 82). Hence the multiplicative inverse of 11 modulo 82 is
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