回复：1726

you are solving X = 30 (mod 32) and X = 44 (mod 58).

Straight forward way to solve this is:

Let X=2Y, then

Y=16x + 15 = 29y + 22;
 
mod 16, we have 13y = 9(mod 16);

Again let 13y = 9 + 16z;

mod 13, we have 3z = 4(mod 13);

Let 3z = 13p + 4;

mod 3, we have p= 2(mod 3);

so p = 3q + 2;

3z = 13p + 4 => z = 13q + 10;

13y = 9 + 16z => y = 13 + 16q ;

X= 58y + 44 => X = 58(13 + 16q) + 44,

substitute q = 1, X = 1726.

所有跟帖： 

• substitute q = 0, and 2 we have two more. -calligraphy- ♂ (0 bytes) () 04/16/2014 postreply 19:35:19

• 大谢如此详细的解释！ -风轻云淡水一方- ♀ (0 bytes) () 04/16/2014 postreply 19:36:46