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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(4): Solution ------ can you check? | Comment 5 of 8 |
(In reply to re(3): Solution ------ can you check? by pcbouhid)

From system of eqns. (II) in the original post , it follows that:

(ab+5107(a-b) + 4942) (mod L)=0, where:

L = LCM(5107, 5108, 5109, 5110, 269) = 91,600,075,916,256,180

or, ab + 5107a - 5107b (mod L)= -4942

or, ab + 5107a - 5107b - 5107^2 (mod L)= -4942 - 5107^2 = -26086391

or, (a-5107)(b+5107) (mod L) = -26086391

So, the  rhs seems to be negative rather than positive.


  Posted by K Sengupta on 2008-08-01 13:28:02
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