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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d052c87-270c-45f0-8400-be0ac3e8daa5%2Fce104582-177d-4cda-8ff6-3ba0c3a8b441%2Fzdke3t_processed.png&w=3840&q=75)
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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