Find the complete solution to B: n ≡ 5 (mod 12) and n ≡ 11 (mod 18)

Question

asnaseer · Answer

$$n\equiv5(mod 12)$$implies n is some multiple of 12 with 5 remainder:$$n=12a+5$$
$$n\equiv11(mod18)$$implies n is also some multiple of 18 with remainder 11:$$n=18b+11$$so n has to satisfy both conditions:$$n=12a+5=18b+11$$$$12a-18b=6$$$$2a-3b=1$$$$a=(1+3b)/2$$a and b must be integers, so (1+3b) must be even which means b must be odd. this gives you:$$b = 1, 3, 5, 7, ...$$$$n=29,65,101,137,...$$
this can be summarised as:$$n=29+36m$$$$m=0,1,2,3,...,\infty$$

anonymous · Answer

Awesome, now, what would be a general rule for solving  a system of linear congruences when the chinese remainder theorem does not apply... i.e gcd(m1, m2) does not equal 1. So, the system has a solution iff (something) and if n is a solution the complete

anonymous · Answer

solution is (something)

anonymous · Answer

If it helps, im referring to how the A: n ≡ 2(mod 12) and n ≡ 10(mod 18) does not have a solution