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I have to calculate $2019x \equiv18^{2019} \, mod \, 45$

This is equivalent to: \begin{cases} 2019x \equiv 18^{2019} \, \text{mod} \, 5 \\ 2019x \equiv 18^{2019} \, \text{mod} \, 9 \end{cases} We know that $2019x \, mod \, 5 = 4x $ and $2019x \, mod \, 9 = 3x $. So we get: \begin{cases} 4x \equiv 1 \, \text{mod} \, 5 \\ 3x \equiv 0 \, \text{mod} \, 9 \end{cases}. We can now multiply by the inverses. So we get: \begin{cases} x \equiv 4 \, \text{mod} \, 5 \\ x \equiv 0 \, \text{mod} \, 3 \end{cases}. The last step will be here to use the Chinese Remainder theorem. I'm asking if my method is correct because it is the first time I'm practicing this kind of questions and I want to be sure.

Thanks in advance!

Kabouter9
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  • I think the congruence mod $5$ starts with a wrong right hand side, and the congruence mod $9$ is a little tricky because $9$ is not a prime (but it is a power of a prime). Check over your work. – hardmath Aug 18 '19 at 17:31
  • Thought $3x \equiv 0 mod 9$ iff $3x = 9q$ with q an integer so $x = 3q$. – Kabouter9 Aug 18 '19 at 17:33
  • Yes, that is correct. It will just be a little tricky piecing things back to get the solutions mod $45$. So one way to think about combining the congruences mod $5$ and $9$ is to list all three of the mod $9$ possibilities. – hardmath Aug 18 '19 at 17:37
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    Looks good to me... using CRT you will get one unique solution in mod $15$. Then you can add multiples of $15$ to that to get unique solutions in mod $45$. – AgentS Aug 18 '19 at 17:43
  • How did you get $18^{2019}\equiv 1 \pmod 5$? For that matter how did you get $18^{2019}\equiv 0 \pmod 9$. – fleablood Aug 18 '19 at 21:23
  • Oh... and $3$ does not have an inverse $\mod 9$.... and how did you drop it from $9$ to $3$. – fleablood Aug 18 '19 at 21:27
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    @ganeshie8 Isn't "unique solution*s*" an oxymoron? – fleablood Aug 18 '19 at 21:30
  • @fleablood OP did do it correctly by cancelling $3$ - see the comment above. – Bill Dubuque Aug 18 '19 at 21:31
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    I didn't say the OP didn't do it correctly. I asked the OP to tell us what he was doing and why. – fleablood Aug 18 '19 at 21:34
  • Lol @fleablood XD that occurred to me while typing haha Then the phrase "3 unique solutions" must be a punishable crime :P – AgentS Aug 18 '19 at 21:46
  • What is the abbrevation for OP? – Kabouter9 Aug 19 '19 at 07:25
  • @Kabouter9 OP = Original Post(er), i.e. the initial post (or its author) – Bill Dubuque Aug 19 '19 at 11:46

2 Answers2

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The method is correct but it should be $\!\bmod 5\!:\ {-}x\equiv 4x\equiv \color{#c00}2,\,$ so $\,x\equiv -2\equiv 3,\,$ and also $\,x\equiv 0\equiv 3\pmod{\!3} \iff x\equiv 3\pmod{\!15}\,$ by LCM or CCRT = Constant case CRT.

Bill Dubuque
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$18^{2019} \equiv (-2)(-2)^{2018} \equiv (-2)(-1)^{1009} \equiv 2 \text{ (mod 5)}$

$2019 x \equiv -x \equiv 2 \Rightarrow x \equiv 3 \text{ (mod 5)}$

Luckily, the second congruence is $x \equiv 0 \text{ (mod 3)}$. No need for CRT, since 3|3

Combine both congruences: $$x \equiv 3 \text{ (mod 15)}$$

albert chan
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