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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 \(n\equiv 2\pmod 3\)

In general :
\(a\equiv b\pmod{n}\) and \(c\equiv d\pmod{n}\) implies \(ac\equiv bd \pmod{n}\)

first think to reduce 11 .-.
in order of 3 set of residues are {0,1,2}

well, yes but how do we go from 325n ≡ n (mod 3) to n ≡ 2 (mod 3) using 11 ≡ 2 (mod 3)

|dw:1438392886462:dw|

ok, so 325n ≡ 11 (mod 3) and 11 ≡ 2 (mod 3), by transitivity, we have
325n ≡ 2 (mod 3)

|dw:1438392895131:dw|

|dw:1438392929478:dw|

:O ohhhhhhhh.... I see it now. thank you @ganeshie8 @jim_thompson5910

notice that we're also using commutative property
\(a\equiv b\pmod{n} \implies b\equiv a\pmod{n}\)