perplexus.info :: Just Math : A system of modular equations

A system of modular equations (Posted on 2008-07-20)

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

Submitted by pcbouhid

Rating: 5.0000 (1 votes)

Solution: (Hide)

KS´comments present the right solution to this problem, which involves the evaluation of the modular inverse. A method for this evaluation can be seen in the submitter comment (the last one).

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re(7): Solution ------ can you check?	K Sengupta	2008-08-05 11:57:01
re(6): Solution ------ can you check?	pcbouhid	2008-08-04 09:11:30
re(5): Solution ------ can you check?	K Sengupta	2008-08-02 13:07:24
re(4): Solution ------ can you check?	K Sengupta	2008-08-01 13:28:02
re(3): Solution ------ can you check?	pcbouhid	2008-08-01 13:05:37
re(2): Solution ------ can you check?	K Sengupta	2008-08-01 06:32:41
re: Solution ------ can you check?	pcbouhid	2008-07-31 17:12:43
Solution	K Sengupta	2008-07-20 14:26:13

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info