Modulo Problem

[Trenters4325]

So then we have \(\displaystyle 19x=141+180n.\) Take this mod 19 and we get \(\displaystyle 9n+8\equiv 0 mod19\)or \(\displaystyle 9n\equiv 11\equiv 144mod 19.\) Divide both sides by 9 to get\(\displaystyle n\equiv 16mod 19.\) So we take\(\displaystyle n=16.\)So then\(\displaystyle 19x=141+180 *16\Rightarrow x=159.\)

I saw this explanation somewhere and I am having trouble figuring out how they went from the first thing in latex to the second. It seems like they took the mod19 of both sides and then did something else.

 

[galactus]

19(9n)=171n, 180n-171n=9n

19(7)=133, 141-133=8

Therefore, \(\displaystyle 9n+8\equiv\0 mod 19\)

That's the best I can do with it.

 

[Trenters4325]

So, did they take mod19 of both sides?
The mod19 of the left side is clearly zero.
And the right side...

 

[pka]

So, did they take mod19 of both sides?
The mod19 of the left side is clearly zero.
And the right side...

You must post the exact statement of any problem you are asking about.
Just posting a solution does not tell us what the problem is.
Particularly in working with modulo arithmetic, there are many different approaches to any problem. I read your post and I could no possibly answer your question.
I have no idea what is going on in that solution.

 

[Trenters4325]

Here's the original problem:

Find the smallest positive integer solution to \(\displaystyle $\tan 19x^\circ=\frac{\cos 96^\circ+\sin 96^\circ}{\cos 96^\circ-\sin 96^\circ}.$\)

After some manipulation, we get \(\displaystyle $\frac{1+\tan{96}}{1-\tan{96}}=\tan{(45+96)}$\) and \(\displaystyle $\tan{141}=\tan{19x}$\), hence the equation from my initial question. I understand everything except the modular arithmatic used to produce that second latexed equation. In particular, I do not understand where the 0mod19 comes from.

 

[Trenters4325]

Should I post this in a different forum?