I am practising solving congruence and have found that my method doesn't always give me the lowest possible answer. I assume the process below would find the lowest possible answer, but I checked and 2 is the lowest. What step am I missing in this process:
3(2x)≡3(1)(mod 9)
2x≡1 (mod 9)
The closes value to 0 by adding/subtracting 9 to 1 is -8
2x≡-8 (mod 10)
Then, we divide by 2:
x≡-4 (mod 9)
Finally, since we cannot have a negative, we need to keep adding by 9 until we get a positive integer:
x≡5 (mod 9)
This only gives us one solution, and there are 3 in total. We keep adding three to give us our other two answers:
x≡5 (mod 9)
x≡8 (mod 9)
x≡11 (mod 9)
How would I have known that it starts at 2 and not 5?