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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9134c255-a2be-4530-8da1-a6d66778d97d%2F04f4ed92-65a6-4954-95db-b99355f4c1ae%2Fg5i9x1n_processed.jpeg&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 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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