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Getting an X for Chinese Remainder Theorem (CRT)

In the "Easy CRT" part of the answer to this problem, the author demonstrates that (-3/77) mod 65 is equal to 16. I don't understand - how is this accurate? I sort of understand the steps, but wouldn't the answer just be 62/77?

Thanks, and I apologize if I've missed something obvious!

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The definition of $\frac{1}{x}$ is that $\frac{1}{x}$ is the quantity such that $x \cdot \frac1x = 1$ (which may or may not exist). Therefore $$ \frac{-3}{77} \equiv 16 \pmod {65} \text{ if and only if } -3 \equiv 77 \cdot 16 \pmod {65} $$ This happens if and only if $$77 \cdot 16 + 3\equiv 0 \pmod {65}$$ which by definition of congruence mod $65$ says that $$65 \mid (77 \cdot 16 + 3)$$ which is true because $$ 77 \cdot 16 + 3 = 1235 = 65 \cdot 19. $$

Trevor Gunn
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In the ring of integers mod. $n$, there is no ‘real’ fraction. Some elements have reciprocals, others don't, depending whether they're coprime with the modulus or not.

For instance, modulo $14$, $5$ is a unit, since $3\cdot 5\equiv 1\pmod{14}$, and $5^{-1}=3$, so one can be tempted to write, say, $\dfrac 45$ instead of $4\cdot 5^{-1}=4\cdot 3=12\pmod{14}$. This way of writing is very confusional, from my point of view.

In the present case, we have $77\equiv 12\pmod{65}$ and one computes with the extended Euclidean algorithm that $12^{-1}\equiv -27$, so that $$-\frac 3{77}=(-3)(-27)=81\equiv 16\pmod{65}.$$

Bernard
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