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I am doing simple question on congruence modulo arithmetic, and my attempt is:

$77546 + x - 465 = 513* 663 \pmod{92}$
=> $77081 + x = (460+53)*(644+19) \pmod{92}$
=> $ (92000-14919) + x = (53)*(19) \pmod{92}$
=> $ (92000-(9200+5719)) + x = (-39)*(19) \pmod{92} $
=> $ (92000-(9200+4600+920+184+15))+ x =(-40+1)*(19) \pmod{92} $
=> $ -15 + x =(-760+19)\pmod{92} $
=> $ -15 + x =-741\pmod{92} $
=> $ -15 + x =(-736 -5)\pmod{92} $
=> $ 77+ x=-5 \pmod{92}$
=> $77+ x= 87 \pmod{92}$
=> $x = 10\pmod{92}.$

The minimum positive value (as need positive values only) is $x = 10$. It is correct, as per the answer.

But, I feel confused over the value $10 \pmod{92} $ that can be added. Can I add any value belonging to the set : $10+ 92n, \forall n \in \mathbb{Z+}$.

The next question is only about repeating the correct formatting for a new question, with nothing to ask:

$2241- 43*7572 \pmod{94}$
=> $(1880 + 376 -15) -43*(9400-1880 + 52) \pmod{94}$
=> $-15 -43*52 \pmod{94}$
=> $-15 -(40+3)*(50+2) \pmod{94}$
=> $-15 -(2000+80 +150 +6) \pmod{94}$
=> $-15 -(2000+236) \pmod{94}$
=> $-15 -(2000+(188+48)) \pmod{94}$
=> $-15 -2048 \pmod{94}$
=> $-15 -74 \pmod{94}$
=> $-89 \pmod{94}$
=> $5 \pmod{94}$

jiten
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    This is unreadable. Please rewrite the question for clarity. – lulu Jan 15 '18 at 20:53
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    This is completely unclear. Your title suggests that you are attempting to show that a certain equality is true, but the work you show does not lead to a conclusion of an equality, and the words following your confusing work do not help at all. What is the question, and where is the work that you think provides an answer? – Steve Kass Jan 15 '18 at 20:58
  • @lulu It is readable hopefully. – jiten Jan 15 '18 at 20:59
  • @SteveKass It is readable hopefully. – jiten Jan 15 '18 at 21:00
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    It is still unreadable. As you can clearly see from the comments, nobody can sort out what $77546+\cdots -465$ means, and the rest just looks like a hopeless mess. – lulu Jan 15 '18 at 21:01
  • @lulu Sorry, I copied the source verbatim. It is now okay, I hope. – jiten Jan 15 '18 at 21:04
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    Ok, at least now the question makes sense. Your work is still hard to follow...you just string together equivalences without regard for whether the terms are equivalent or not. But your final answer looks correct. The answer is $x\equiv 10\pmod {92}$ which is the same as saying that $x$ could be any integer of the form $10+92n$ for $n\in \mathbb Z$. – lulu Jan 15 '18 at 21:10
  • I hope you must have meant for before the final version of OP, that I have strung equivalences together without them being equivalent. If the error is still there, please tell me, as I am unable to notice now. – jiten Jan 15 '18 at 21:13
  • @lulu so, your answer is I can add $10 -92$ also, i.e. not restricted to positive integers alone. – jiten Jan 15 '18 at 21:14
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    It's a question of what you are looking for. If you want all integers $x$ that make the congruence true then you can certainly get negative integers as well as positive integers. If you just want the residue class that makes it true, then the answer is $x\equiv 10 \pmod {92}$. – lulu Jan 15 '18 at 21:16
  • @lulu Thanks. Please also respond to my earlier comment regarding the logical error introduced by formatting, if still present. I hope it is all correct now. – jiten Jan 15 '18 at 21:17
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    The biggest part of what was unclear is that you never made it clear that you were attempting to do is solve an equation for $x$. I think the comments answer your question now. – Steve Kass Jan 15 '18 at 21:20
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    The writing is still very unclear. $\equiv$ is the standard notation for congruence. You use $=$ instead, but then you use $\equiv$ apparently to mean $\iff$ which is non-standard. – lulu Jan 15 '18 at 21:21
  • @lulu Very correct, I have used in that sense when linking one set of equalities to another. – jiten Jan 15 '18 at 21:23
  • @lulu I hope that error of using $\equiv$ for $\iff$ is not any more. – jiten Jan 15 '18 at 21:27
  • @lulu sorry for troubling, but I hope this final version is without errors, and can use it in future too. – jiten Jan 15 '18 at 21:36

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