All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
A system of modular equations (Posted on 2008-07-20) Difficulty: 3 of 5
Solve the following set of equations for positive integers a and b:

1919(a)(b) mod 5107 = 1
1919(a+1)(b-1) mod 5108 = 5047
1919(a+2)(b-2) mod 5109 = 1148
1919(a+3)(b-3) mod 5110 = 3631
1919(a+4)(b-4) mod 5111 = 2280

See The Solution Submitted by pcbouhid    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(5): Solution ------ can you check? | Comment 6 of 8 |
(In reply to re(4): Solution ------ can you check? by K Sengupta)

If the rhs is positive, that is:

(a-5107)(b+5107) (mod L) = 26086391 = 4241*6151 = 6151*4241 = 1* 26086391 = 26086391*1

Then, (a, b) = (9348, 1044), (5108, 26081284)

For, (a, b) = (9348, 1044),  we have:
1919*(9348)*(1044) (mod 5107)= 5106, so the first equation is not satisfied for these values. Indeed, the other four equations are not satisfied for these values.

Similarly, for (a, b) = (5108, 26081284), none of the five given equations are satisfied.

However, if the rhs is negative, that is (a-5107)(b+5107) (mod L) = -26086391, then:
(a, b) = (866, 1044), (5106, 26081284), as was demonstrated in  my original post.

It may be verified that each of (866, 1044) and (5106, 26081284), satisfies the five given equations.

Edited on August 2, 2008, 1:10 pm
  Posted by K Sengupta on 2008-08-02 13:07:24

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information