i learnt about the Chinese remainder theorem, and im trying to solve the following question:
find the minimal solution x for
(1)x = 11 mod 24
(2)x = 5 mod 18
(3)x = 5 mod 30
i know that in order to use the Chinese reminder theorem i need the modulos to be distinct
how can i solve it regardless?
i tried the following(and i think this is good):
24 = 8*3
18 = 2*9
30 = 2 * 3 * 5
so(1) is the same as:
(1*)x = 11 = 3 mod 8
(2*)x = 11 = 2 mod 3
(2) is the same as:
(3*)x = 5 = 1 mod 2
(4*)x = 5 mod 9
(3) is the same as:
(5*)x = 5 = 1 mod 2
(6*)x = 5 = 2 mod 3
(7*)x = 5 = 0 mod 5
from 1* we get x = 8k+3 so it also means x = 1 mod 2 so if x satisfies 2* it also satisfies 3*,5*)
from 4* we get x = 9k + 5 and this implies x = 2 mod 3, so if x satisfies 4* it also satisfies 2*,6*
so it is enough to solve :
x = 3 mod 8 x = 5 mod 9 x = 0 mod 5
and from here i can use the Chinese remainder theorem
this way of solving is fine right? is there a better way? is this a good way to approach solving such equations when i cant use the chinese remainder theorem because the modulus are not distinct?
thanks!!