Try to find all integers x whose remainder is, 3, and 9, 8, and 7 respectively.

Introduction to algorithms exercise 31.5-2

XLimit 1 (mod 9)

XLimit 2 (mod 8)

XRelease 3 (mod 7)

 

According to the remainder theorem of China,

N = 504

A1 = 1,N1 = 9,M1 = 56,56-1 limit 5 (mod 9)

A2 = 2,N2 = 8,M2 = 63,63-1 limit 7 (mod 8)

A3 = 3,N3 = 7,M2 =-1 limit 4 (mod 7)

C1 = 56 (5 mod 9) = 280

C2 = 63 (7 mod 8) = 441

C3 = 72 (4 mod 7) = 288

 

A	Bytes	1. 280 + 2. 441	+ 3 · 288 (mod 504)
	Bytes	280 + 882 + 864	(Mod 1, 504)
	Bytes	10	(Mod 1, 504)

 

Therefore, all the solutions that make the equation come true at the same time are integers in the form of 10 + 504 K (K is an arbitrary integer.