Using the Chinese Remainder Theorem, solve:
x ≡ 1 mod 2
x ≡ 2 mod 3
x ≡ 3 mod 5
x ≡ 4 mod 11
GCF(2,11) = 1
GCF(3,11) = 1
GCF(5,11) = 1
Since all 6 GCF calculations equal 1
the ni's are pairwise coprime
We can use the regular CRT Formula
Take the product of each ni
N = n1 x n2 x n3 x n4
N = 2 x 3 x 5 x 11
N = 330
ci = | N |
ni |
c1 = | 330 |
2 |
c1 = 165
c2 = | 330 |
3 |
c2 = 110
c3 = | 330 |
5 |
c3 = 66
c4 = | 330 |
11 |
c4 = 30
x = a1(c1y1) + a2(c2y2) + a3(c3y3) + a4(c4y4)
x = a1(165y1) + a2(110y2) + a3(66y3) + a4(30y4)
Note: The ai piece is factored out
We will use this below
Using 1 modulus of 2 and c1 = 165
calculate y1 in the equation below:
Using 2 modulus of 3 and c2 = 110
calculate y2 in the equation below:
Using 3 modulus of 5 and c3 = 66
calculate y3 in the equation below:
Using 4 modulus of 11 and c4 = 30
calculate y4 in the equation below:
x = a1(165y1) + a2(110y2) + a3(66y3) + a4(30y4)
x = 1 x 165 x 1 + 2 x 110 x -1 + 3 x 66 x 1 + 4 x 30 x -4
x = 165 - 220 + 198 - 480
x = -337
-337 ≡ 1 mod 2
Add remainder of 1 to -338 = -337
-337 ≡ 2 mod 3
Add remainder of 2 to -339 = -337
-337 ≡ 3 mod 5
Add remainder of 3 to -340 = -337
-337 ≡ 4 mod 11
Add remainder of 4 to -341 = -337