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If $n$ and $m$ are positive integers and $n^{9}=19m+r$, then the possible values for $r$ modulo 19 are.

  1. Only 0
  2. Only 0, $\pm$ 1
  3. Only $\pm$ 1
  4. None of the above
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According to Fermat's little theorem, if $p$ is a prime number, then for any integer $n$,

$$n^{p-1}\mod p = 1$$

In question, $p$ is 19, and so for any integer $n$, $(n^{18})\mod 19 = 1$.

So for $n^9$, we find square roots of 1 mod 19, which is $\pm 1$.
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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^(p−1) 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

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