Try this problem of TIFR GS-2010 using your concepts of number theory and congruence based on Positive Integers.
If $n$ and $m$ are positive integers and $n^{19}=19m+r$ then the possible values for $r$ modulo $19$ are:-
NUMBER THEORY
CONGRUENCE
FERMAT'S LITTLE THEOREM
Answer:only $0$,$\pm1$
TIFR 2010|PART B |PROBLEM 12
ELEMENTARY NUMBER THEORY DAVID M.BURTON
$r(mod 19)=(n^9-19m)(mod 19)=n^9(mod 19)-19m(mod 19)=n^9(mod 19)$ [because $19\mid19m$]
Now there can be two cases
casei) $19\mid n$
caseii) n is not divisible by 19
If $19\mid n$ then $19\mid n^9$ resulting in $n^9\equiv 0(mod 19)$
But if $n$ is not divisible by $19$ then according to Fermat's little theorem,
$n^{19-1}\equiv1(mod 19) =n^{18}\equiv 1(mod 19) = 19\mid n^{18}-1 =19\mid[n^9+1][n^9-1]$
i.e If $19\mid n^9+1$ then $n^9\equiv(-1)(mod 19) =r\equiv(-1)(mod 19)$
If $19\mid n^9-1$ then $n^9\equiv1(mod 19) = r\equiv1(mod 19)$
Thus $r(mod 19)=0$,$\pm1$

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