(a) Find an inverse for 73 modulo 660. First use the extended Euclidean algorithm to find the greatest common divisor of 660 and 73 and express it as a linear combination of 660 and 73. Step 1: Find q, and r, so that 660 - 73 -4, + where Osr, <73. Then r,- 660 - 73 - 4,- Step 2: Find q, and r, so that 73 -9+ where 0sr Then r,-73- Step 3: Find and so that 1-24+ where osr <- Then, 4, -0 Step 4: Select the comrect statement from the options below. O Becauser-0, god (660, 73) - 73 -r,4,-1. O Because r,- 0, god (660, 73) -- - O Becauser- 1, god (660, 73) - 73 - 2-0. O Because r - 1, god (660, 73)--73-4-0. O Becauser- 1, god (660, 73) -- -0. Next, substitute numerical values backward through the preceding steps, simplifying the results for each step, to find god (660, 73) -1- 660s + 73r, where s- and t- It follows that 73 - (0 =1 (mod 660), and so |is an inverse for 73 modulo 660. (b) Find the least positive solution for the congruence: 73x = 125 (mod 660). Step 1: By part (a), 73 0) =1 (mad 660). Multiply both sides by 125 to obtain 73 - 125 =1- 125 (mad 660) = 125 (mod 660). Let x- 125- Then 73x = 125 (mod 660). Nate that x is a solution to the given congruence, but it may not be the least positive solution. Step 2 Let v be the remainder abtained when x is divided by 660. Then v- Jand, by Theorem 8.4.1 and Theorem 8.4.3, 73x = 73v (mod 660). Hence, v is also a solution to the congruence. In other words, 73v = 125 (mod 660). Suppose u is any positive solution to the congruence that is less than or equal to v. Then 73- u= 73 - J(mad 660). Since 73 is prime, gdc(73, 660) - 1. Thus, Selec- can be applied to conclude that u- Therefore, is the least positive solution to the congruence.
(a) Find an inverse for 73 modulo 660. First use the extended Euclidean algorithm to find the greatest common divisor of 660 and 73 and express it as a linear combination of 660 and 73. Step 1: Find q, and r, so that 660 - 73 -4, + where Osr, <73. Then r,- 660 - 73 - 4,- Step 2: Find q, and r, so that 73 -9+ where 0sr Then r,-73- Step 3: Find and so that 1-24+ where osr <- Then, 4, -0 Step 4: Select the comrect statement from the options below. O Becauser-0, god (660, 73) - 73 -r,4,-1. O Because r,- 0, god (660, 73) -- - O Becauser- 1, god (660, 73) - 73 - 2-0. O Because r - 1, god (660, 73)--73-4-0. O Becauser- 1, god (660, 73) -- -0. Next, substitute numerical values backward through the preceding steps, simplifying the results for each step, to find god (660, 73) -1- 660s + 73r, where s- and t- It follows that 73 - (0 =1 (mod 660), and so |is an inverse for 73 modulo 660. (b) Find the least positive solution for the congruence: 73x = 125 (mod 660). Step 1: By part (a), 73 0) =1 (mad 660). Multiply both sides by 125 to obtain 73 - 125 =1- 125 (mad 660) = 125 (mod 660). Let x- 125- Then 73x = 125 (mod 660). Nate that x is a solution to the given congruence, but it may not be the least positive solution. Step 2 Let v be the remainder abtained when x is divided by 660. Then v- Jand, by Theorem 8.4.1 and Theorem 8.4.3, 73x = 73v (mod 660). Hence, v is also a solution to the congruence. In other words, 73v = 125 (mod 660). Suppose u is any positive solution to the congruence that is less than or equal to v. Then 73- u= 73 - J(mad 660). Since 73 is prime, gdc(73, 660) - 1. Thus, Selec- can be applied to conclude that u- Therefore, is the least positive solution to the congruence.
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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![Certainly! Below is a transcription of the text from the image with explanations that would fit an educational website.
---
### Finding an Inverse for 73 modulo 660
#### (a)
**Objective:** Find an inverse for 73 modulo 660.
1. **First Step:** Use the extended Euclidean algorithm to find the greatest common divisor (gcd) of 660 and 73, expressing it as a linear combination of 660 and 73.
- **Step 1:** Find \( q_1 \) and \( r_1 \) such that:
\[
660 = 73 \times q_1 + r_1, \quad \text{where } 0 \leq r_1 < 73.
\]
Then, \( r_1 = 660 - 73 \times q_1 \).
- **Step 2:** Find \( q_2 \) and \( r_2 \) so that:
\[
73 = r_1 \times q_2 + r_2, \quad \text{where } 0 \leq r_2 < r_1.
\]
Then, \( r_2 = 73 - r_1 \times q_2 = \_\_\_\_\_\_ \).
- **Step 3:** Find \( q_3 \) and \( r_3 \) so that:
\[
r_1 = r_2 \times q_3 + r_3, \quad \text{where } 0 \leq r_3 < r_2.
\]
Then, \( r_3 = \_\_\_\_\_ - (\_\_\_\_\_\_\_) \times q_3 = \_\_\_\_\_\_ \).
2. **Step 4:** Select the correct statement from the options below.
- Because \( r_3 = 0 \), gcd (660, 73) = \( r_2 = r_1 - q_3 \times q_2 = 1 \).
- ...
Next, substitute numerical values backward through the preceding steps, simplifying the results for each step to find gcd (660, 73) = 1 = 660s + 73t, where \( s = \_\_\_\_\_\_ \) and \( t = \_\_\_\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc357075-d330-4116-93bc-b2819b26405c%2Fdbddd2d9-8942-49d1-be01-528f5c4c285d%2Fu4vd2ps_processed.png&w=3840&q=75)
Transcribed Image Text:Certainly! Below is a transcription of the text from the image with explanations that would fit an educational website.
---
### Finding an Inverse for 73 modulo 660
#### (a)
**Objective:** Find an inverse for 73 modulo 660.
1. **First Step:** Use the extended Euclidean algorithm to find the greatest common divisor (gcd) of 660 and 73, expressing it as a linear combination of 660 and 73.
- **Step 1:** Find \( q_1 \) and \( r_1 \) such that:
\[
660 = 73 \times q_1 + r_1, \quad \text{where } 0 \leq r_1 < 73.
\]
Then, \( r_1 = 660 - 73 \times q_1 \).
- **Step 2:** Find \( q_2 \) and \( r_2 \) so that:
\[
73 = r_1 \times q_2 + r_2, \quad \text{where } 0 \leq r_2 < r_1.
\]
Then, \( r_2 = 73 - r_1 \times q_2 = \_\_\_\_\_\_ \).
- **Step 3:** Find \( q_3 \) and \( r_3 \) so that:
\[
r_1 = r_2 \times q_3 + r_3, \quad \text{where } 0 \leq r_3 < r_2.
\]
Then, \( r_3 = \_\_\_\_\_ - (\_\_\_\_\_\_\_) \times q_3 = \_\_\_\_\_\_ \).
2. **Step 4:** Select the correct statement from the options below.
- Because \( r_3 = 0 \), gcd (660, 73) = \( r_2 = r_1 - q_3 \times q_2 = 1 \).
- ...
Next, substitute numerical values backward through the preceding steps, simplifying the results for each step to find gcd (660, 73) = 1 = 660s + 73t, where \( s = \_\_\_\_\_\_ \) and \( t = \_\_\_\
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