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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    
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re(7): Solution ------ can you check? Comment 8 of 8 |
(In reply to re(6): Solution ------ can you check? by pcbouhid)

Thanks  PCB. for the acknowledgement

At the time this problem was published, my proficiency with the
Euclidean algorithm  was rather rusty, and the modular inverse for the very first equation that I came up with proved to be erroneous upon due verification.

Apart from the final equation, the M-inverse of which is trivial, for the other four equations I had to resort to a combination of brute force and Chinese Remainder Theorem. 

Edited on August 5, 2008, 4:45 pm
  Posted by K Sengupta on 2008-08-05 11:57:01

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