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3 Answers

Best answer
27 votes
27 votes

Fermat's Little Theorem :

$$a^p ≡ a (\mod p)$$                        
According to Modular Arithmetic    $a ≡ b (\mod n)$ if their difference $(a-b)$ is an integral multiple of $n$ $( n$ divides $(a-b) )$

So, $( a^p - a )$ is an integral multiple of  $p.$ Now as $a$ is not divisible by $p$ so definitely,  $( a^{p-1} -1)$ is an integral multiple of $p.$ This simply means if we divide $a^{p-1}$  by $p,$ the remainder would be $1$. i.e., $ a^{p-1} \mod p = 1.$ 

Put the values in the formula.
$p=17$
So, $p-1 =16.$

Correct Answer: $D$

edited by
6 votes
6 votes
D) fermat's little theorem
3 votes
3 votes

Using Fermats Little Theorem

p: prime 
a : integer Not prime 
then
ap-1 mod p is always 1 

Here p : 7  Hence p-1  is 16

Answer:

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