0

So I'm not very good at the chinese remainder theorem, but I think that x being 551 works. But I'm not sure how the negative 1 affects the equation. Also, if x=551 works, then would you multiply 8×9×5=360 and then subtract 360 from 551 to get x= 191 (mod 360)? Any help on this would be great! Thank you!

SarahCh
  • 13
  • 2
  • $551\equiv 1\pmod 5$, not $-1$. If you dislike the $-1$ for whatever reason you can just replace it with $4$. That is to say, $x\equiv -1\pmod 5\iff x\equiv 4 \pmod 5$. But, really, there's nothing wrong with $-1$. – lulu Apr 08 '20 at 14:10
  • Thanks! That helps a lot! When I get the new number (let's say its Y) would I still multiply the (8×9×5) and subtract it from Y to write the answer in mod form? – SarahCh Apr 08 '20 at 14:13
  • Please use mathjax to format your text. If you put the mous over a formula an open the context menu by pressing the right mouse button you can select "Show Math As => TeX Commands" to display the Latex of a formula – miracle173 Apr 08 '20 at 14:15
  • To your question: once you have some $x_0$ that works then the general solution is indeed $x_0+360$ since $360$ is the least common multiple of $8,9,5$. – lulu Apr 08 '20 at 14:18
  • You can think of $-1\pmod 5$ as $4 \pmod 5$ if that makes anything easier for you but you could also that of it as anything else (such as $19\pmod 5$) if you want. Negatives don't change anything. ... $551$ doesn't work as $551 \equiv 1 \not \equiv -1 \pmod 5$. (Whatever does work must end in a $4$ or a $9$ if it is positive or in a $1$ or a $6$ if it is negative) But yes. If any $x$ works then so does ant $x \pm 360k$ will work – fleablood Apr 08 '20 at 21:49
  • $551$ (and therefore $191$) is congruent to $7\pmod 8$ ($191-7=184$ is divisible by $8$). And congruent to $2\pmod 9$ ($191-2=189$ is divisible by $9$). But $551$ not $191$ is congruent to positive $1\pmod 5$ but not to $-1$ nor to $4$ because $191-(-1) =192$ or $191-4=187$ are neither divisible by $5$. – fleablood Apr 08 '20 at 21:57

1 Answers1

0

Negative numbers follow the same rules as positive in modular equations so we can say $$x\equiv-1\equiv -1 + 5 \equiv 4\pmod{5} $$ The reason we might use them is that we can see that the first equation can be rewritten as $x\equiv -1\pmod{8}$, and then we can realize that $8*5-1 =39$ satisfies both $$8*5-1 \equiv -1 \pmod{5}$$ and $$8*5-1 \equiv -1 \pmod{8}$$ Unfortunately, however, $39 \equiv 3\pmod{9}$, so we're not done yet. We have to add a multiple of $40 \equiv 4\pmod{9}$ in order to make this number also $2\pmod{9}$. To do this, we must solve the equation $$39 + 40n \equiv 3 + 4n \equiv 2 \pmod{9}$$ Then we can rearrange this as $$4n \equiv 3-2 \equiv 8\pmod{9}$$ And see that $n=2$ solves it. Thus an answer is $39 + 40*2 = 119$. And as $5,8,9$ are all relatively prime, the solutions will be precisely $119 + 5*8*9k$ for $k\in \mathbb{Z}$.

Isaac Browne
  • 8,076