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I've marked this up the best way I can:

$0 \equiv (19+16x) \pmod{15-x}$

I can repeat this equation filling in $x$, which gets increased by one with each pass. When you get to $x$ = 8, the remainder is finally $0$. My goal is to solve this equation to get straight to $x$ = 8 without having to do the passes.

To me that equation says solve for $x$ to make the remainder $= 0$. If this is correctly written then I've been trying to figure out how to solve this. My algebra is limited and I've done research about Chinese Remainder Theorem and Euclid's algorithm but I'm not sure if I have this set up right.

Any help would be highly appreciated.

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  • $\begingroup$ What is $y$? There are only $x$'s. $\endgroup$ Commented Jan 17, 2015 at 23:49
  • $\begingroup$ My apologies. I was editing and that was a typo. It's now fixed. $\endgroup$ Commented Jan 17, 2015 at 23:54

2 Answers 2

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We are working modulo $15-x$ and are trying to solve $0 \equiv 19+16x$. Well, we can add $16(15-x)$ to both sides and get $0 \equiv 16(15-x) \equiv 16 \cdot 15 +19 \equiv 259 \mod 15-x$. So we get the result that $259=(15-x)k$ for some $k\mathbb \in Z$.

$259=37 \cdot 7$. We must have that $15-x$ divides $259$ and by the above factorization, the divisors are $ \pm 1, \pm 7, \pm 37$ and $\pm 259$. So set $15-x$ equal to all of these numbers to get all different possible values for $x$.

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  • $\begingroup$ That's making more sense. I set 15 - $x$ equal to those numbers and never got $x$ = 8, which is the answer I'm looking for. I also got fractions for the result on some. I re-edited the question to hopefully clear things up. Sorry if I'm simply misunderstanding. $\endgroup$ Commented Jan 18, 2015 at 3:25
  • $\begingroup$ The answer should read $...16\cdot 15+19=259$, not $255$. $\endgroup$ Commented Jan 18, 2015 at 4:23
  • $\begingroup$ Even with the corrected 259 I'm not seeing how you'd get $x$ = 8, which is the answer I'm looking for. $\endgroup$ Commented Jan 18, 2015 at 5:11
  • $\begingroup$ @Grant you don’t have set $x$ equal to these values, you have to set $15 - x$ equal to these values, and, you will get $x=8$ when $15-x = 7$. Do you get it now? $\endgroup$ Commented Jun 20, 2019 at 10:33
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$$0 \equiv 19+16x\equiv 259 \pmod{15-x}\quad\overset{259=7\cdot37}{\implies}$$

$$15-x=(\pm1,\pm7,\pm37,\pm259) \implies$$

$$x=(-244,-22,8,14,16,22,52,274)$$

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