The book doesn't provide a clear explanation. Find an integer n that satisfies 325n ≡ 11 (mod 3). It just says Since 325 ≡ 1 (mod 3) and 11 ≡ 2 (mod 3), we must find an integer n such that n ≡ 2 (mod 3).

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The book doesn't provide a clear explanation. Find an integer n that satisfies 325n ≡ 11 (mod 3). It just says Since 325 ≡ 1 (mod 3) and 11 ≡ 2 (mod 3), we must find an integer n such that n ≡ 2 (mod 3).

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\(325n=(3\cdot 108+1)n=3\cdot108n+n\equiv 0+n=n\pmod 3\)
similarly \(11=3\cdot3+2\equiv2\pmod 3\)
hence if \(325n\equiv 11\pmod 3\) then \(n\equiv 11\pmod 3\) and ultimately \(n\equiv 2\pmod 3\)

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In general : \(a\equiv b\pmod{n}\) and \(c\equiv d\pmod{n}\) implies \(ac\equiv bd \pmod{n}\)
basically if \(a\equiv b\pmod n\) and \(b\equiv c\pmod n\) then \(a\equiv c\pmod n\), it's transitivity
Because we're dealing with mod 3, we can use this trick: Add up the digits of 325 to get 3+2+5 = 5+5 = 10 so 325 = 10 (mod 3) 10/3 = 3 remainder 1 which is why 325 = 10 = 1 (mod 3) The same happens with 11 1+1 = 2 which is why 11 = 2 (mod 3)
@oldrin.bataku ok, let me think through your answer so 325n = (3 * 108 + 1)n = 3(108)n + n. Ok this I get How did you get 3(108)n + n ≡ 0 + n (mod 3). Did you use the fact that n ≡ n (mod 3)?
first think to reduce 11 .-. in order of 3 set of residues are {0,1,2}
@sourwing well, clearly \(3\cdot k\equiv 0\pmod 3\) so the \(3\cdot108n\) term is congruent to \(0\)
@oldrin.bataku oh, so 3(108n) ≡ 0 (mod 3) and n ≡ n (mod 3). adding the two congruences gives 108n + n ≡ 0 + n (mod 3) 325n ≡ n (mod 3). And how does 11 ≡ 2 (mod 3) come in place?
I guess it has something to do with what @ganeshie8 said earlier. 325 ≡ 1 (mod 3) and n ≡ n (mod 3), then 325n ≡ n (mod 3). But what about 11 ≡ 2 (mod 3)?
\[\Large \frac{11}{3} = 3 \text{ remainder } {\color{red}{2}}\] \[\Large 11 \equiv {\color{red}{2}} \ (\text{mod } 3) \]
11 and 2 leave same remainder when divided by 3, so they are congruent. keep subtracting 3 until you get a nonnegative integer less than 3
well, yes but how do we go from 325n ≡ n (mod 3) to n ≡ 2 (mod 3) using 11 ≡ 2 (mod 3)
use transitivity : basically if \(a\equiv b\pmod n\) and \(b\equiv c\pmod n\) then \(a\equiv c\pmod n\), it's transitivity
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ok, so 325n ≡ 11 (mod 3) and 11 ≡ 2 (mod 3), by transitivity, we have 325n ≡ 2 (mod 3)
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ok, so 325n ≡ 11 (mod 3) and 11 ≡ 2 (mod 3), by transitivity, we have 325n ≡ 2 (mod 3) this looks good, apply transitivity again : 325n ≡ 2 (mod 3) and 325n ≡ n (mod 3), by transitivity, we have n ≡ 2 (mod 3)
:O ohhhhhhhh.... I see it now. thank you @ganeshie8 @jim_thompson5910
notice that we're also using commutative property \(a\equiv b\pmod{n} \implies b\equiv a\pmod{n}\)
So so let me recap all this. Given 325n ≡ 11 (mod 3). Since 325n ≡ n (mod 3), by commutative rule, n ≡ 325n (mod 3) by transitivity n ≡ 11 (mod 3) but then 11 ≡ 2 (mod 3), by transitivity n ≡ 2 (mod 3)
Looks good, fun fact : when divided by \(n\), If \(a\) leaves a remainder of \(a'\) and \(b\) leaves a remainder of \(b'\), then \(a*b\) leaves a remainder of \(a'*b'\).
thank you. I might find that fact useful :) @ganeshie8 The author certainly left too many details out. Would have nice to at least mention 325n ≡ n (mod 3). From here, I might be able to see why 11 ≡ 2 (mod 3) is used.
in the start, it can be helpful to keep referring back to the definition every time a congruence appears in the textbook : \[a\equiv b\pmod{n} \iff n\mid (a-b)\]

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