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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?

Cody Bijeaux
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  • $3(2x)\equiv3(1)\pmod9\iff2x\equiv1\color{blue}{\equiv4}\pmod{\color\red3}$; also, you could see $11-9=2$ – J. W. Tanner Nov 26 '23 at 22:18
  • To obtain an equivalent congruence you need to cancel $:!\color{#c00}3:!$ from the $\rm\color{#c00}{modulus}$ too - see the congruence cancellation rule in the linked dupe. Or, said in equivalent divisibility language $,\color{#0a0}3\cdot \color{#c00}3\mid \color{#c00}3:!(2x!-!1)\iff \color{#0a0}3\mid 2x!-!1,,$ by cancelling $,\color{#c00}3\ \ $ – Bill Dubuque Nov 26 '23 at 22:50

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