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Referring here, I understand how to solve the list of congruences.

However, I was presented this particular example in a book and have one minor question.

It goes: $$x \equiv 1 \pmod 3 \implies x = 1 + 3k_1, k_1\in\mathbb Z.$$ Then plug into second we get, $$ 1 + 3k_1 \equiv 2 \pmod 5 \implies k_1 \equiv 2 \pmod 5.$$ Here is where I don't understand. I do not understand how they go from left to right side of arrow. Here is my way: $$ \begin{align} 1 + 3k_1 \equiv 2 \pmod 5 &\implies 5\mid (1 +3k_1 - 2) \\ &\implies 5 \mid (3k_1 - 1) \\ &\implies 3k_1 \equiv 1\pmod 5. \end{align} $$ Thanks in advance.

Arturo Magidin
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Owen Murphy
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    $1+3k_1\equiv 2\pmod{5}$, subtract $1$ from each side to get $3k_1\equiv 1\pmod{5}$ and then multiply by $3^{-1}$ on each side to get $k_1\equiv 3^{-1}\pmod{5}$. Now... what is $3^{-1}$ in this context? It is not the fraction $\frac{1}{3}$ that you'd be used to in the field of rational numbers or real numbers... it is the number who when multiplied by $3$ results in $1$ in the context of this modular arithmetic ring. Here... $3\times 2 = 6$ which is $1$ more than $5$ so $k_1\equiv 3^{-1}\equiv 2\pmod{5}$ – JMoravitz Mar 02 '22 at 15:49
  • @JMoravitz, Thanks, that makes sense. I understand $3^{-1}$ is the multiplicative inverse of 3 to get the unity element in this particular ring. Thanks again. – Owen Murphy Mar 02 '22 at 15:53
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    As an aside, it was quick to spot that $3^{-1}$ was equivalent to $2$ here by inspection, however for larger examples it may not be reasonable to expect to spot what the inverse is equal to at a glance. In such scenarios the extended euclidean division algorithm and bezout's identity will be helpful. – JMoravitz Mar 02 '22 at 15:55

2 Answers2

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multiply by 2 in both sides you'll get $2+6k_1 \equiv 4 \mod 5 \implies 2 + k_1 \equiv 4 \mod 5 \implies k_1 \equiv 2 \mod 5$

Don Freecs
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Starting from the end of your answer: $$3k_1 \equiv 1\mod{5}$$ we can multiply both sides by 2 to get:

$$6k_1 \equiv 2\mod{5}$$ Since,$$ 5k_1 + k_1 \equiv k_1\mod{5}$$

We know that $$\Rightarrow k_1 \equiv 2\mod{5}$$

maxy
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