option B is correct....

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$n^{9}=19m+r \\ \implies r = n^9-19m \\ \implies r modulo 19 = n^9 modulo 19$

Now we need to find $n^9 modulo 19$

Fermat's little theorem is applicable if that prime number p does not divide n.

Fermat's little theorem : $n^{p-1} mod p =1$ , it is obvious that if p divides n then ($n^{p-1} mod p =0 $)

here two cases:

Case 1. 19 divides n. Then $n^{anything} modulo 19 = 0$

i.e. $n^{9} modulo 19 = 0$

means $r modulo 19 = 0$

Case 2: 19 does not divide n, 19 is prime number also, Therefore we can apply Fermat's Little theorem. (This is already well explained by Happy Mittal sir)

Now we need to find $n^9 modulo 19$

Fermat's little theorem is applicable if that prime number p does not divide n.

Fermat's little theorem : $n^{p-1} mod p =1$ , it is obvious that if p divides n then ($n^{p-1} mod p =0 $)

here two cases:

Case 1. 19 divides n. Then $n^{anything} modulo 19 = 0$

i.e. $n^{9} modulo 19 = 0$

means $r modulo 19 = 0$

Case 2: 19 does not divide n, 19 is prime number also, Therefore we can apply Fermat's Little theorem. (This is already well explained by Happy Mittal sir)

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r=n^9-19m so r modulo 19=(n^9-19m)modulo 19=((n^9modulo 19)-(19m modulo 19))modulo 19=n^9modulo 19 .. now if n is divisible by 19 then n^9modulo 19=0 but if n is not divisible by 19 then accorsing to Fermat's little theorem (n^(pā1) mod p =1 for p being prime number and n being any integer not divisible by p)n^18modulo 19=1. So n^9modulo 19=((n^18modulo 19)^(1/2))modulo 19=1^(1/2) modulo 19 = +1 or -1. So option b