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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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