For which integers c, 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
excercise 4.2 #6
Transcribed Image Text:4.2 Linear Congruences
161
6. For which integers c, 0 <c < 30, does the congruence 12x =c (mod 30) have solutions?
When there are solutions, how many incongruent solutions are there?
7. For which integers c, 0 <c < 1001, does the congruence 154x =c (mod 1001) have solu-
tions? When there are solutions, how many incongruent solutions are there?
8. Find an inverse modulo l13 of each of the following integers.
a) 2
b) 3
c) 5
9. Find an inverse modulo l17 of each of the following integers.
a) 4
b) 5
c) 7
d) 16
10. a) Determine which integers a, where 1< a < 14, have an inverse modulo 14.
b) Find the inverse of each of the integers from part (a) that have an inverse modulo 14.
11. a) Determine which integers a, where 1<a < 30, have an inverse modulo 30.
b) Find the inverse of each of the integers from part (a) that have an inverse modulo 30.
12. Show that if ā is an inverse of a modulo m and b is an inverse of b modulo m, then ā b is an
inverse of ab modulo m.
13. Show that the linear congruence in two variables ax + by = c (mod m), where a, b, c, and
m are integers, m > 0, with d = (a, b, m), has exactly dm incongruent solutions if d | c, and
no solutions otherwise.
14. Find all solutions of each of the following linear congruences in two variables.
c) 6x + 3y = 0 (mod 9)
d) 10x + 5y = 9 (mod 15)
a) 2x + 3y = 1 (mod 7)
b) 2x + 4y = 6 (mod 8)
15. Let p be an odd prime and k a positive integer. Show that the congruence x2 = 1 (mod pk)
has exactly two incongruent solutions, namely, x =±1 (mod p).
16. Show that the congruence x² = 1 (mod 2*) has exactly four incongruent solutions, namely,
x =±l or ±(1+ 2k-1) (mod 2k), when k > 2. Show that when k = 1 there is one solution
and that when k =2 there are two incongruent solutions.
%3D
17. Show that if a and m are relatively prime positive integers such that a < m, then an inverse
of a modulo m can be found using O (log' m) bit operations.
18. Show that if p is an odd prime and a is a positive integer not divisible by p, then the congruence
x² = a (mod p) has either no solution or exactly two incongruent solutions.
Computations and Explorations
1. Find the solutions of 123,456,789x = 9,876,543,210 (mod 10,000,000,001).
2. Find the solutions of 333,333,333x = 87,543,211,376 (mod 967,454,302,211).
3. Find the inverses of 734,342; 499,999; and 1,000,001 modulo 1,533,331.
Programming Projects
* Solve linear congruences using the method given in the text.
* Solve linear congruences using the method given in Exercise 4.
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