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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

### 1 comment

option B is correct....

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 $\pm 1$.

If we take n = 19 (+ve integer) then we are having  r modulo 19 = 0
$n^{9}=19m+r \\ \implies r = n^9-19m \\ \implies r modulo 19 = n^9 modulo 19$
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)
B should be right.

take n = 1,2,19

We cannot do this process . As If$a\equiv b(mod n)$ then $a^k\equiv b^k (mod n)$ for any positive integer $k$

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