Question
Given that p is a prime number. p2[p27−1p] is always divisible by?
Answer: Option D
:
D
According to fermat's theorem " If P is a prime number and N is prime to P, then Np−N is divisible by P."
Open the brackets, p2[p27−1p]=p29−p will always be divisible by 29, as it is prime.
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:
D
According to fermat's theorem " If P is a prime number and N is prime to P, then Np−N is divisible by P."
Open the brackets, p2[p27−1p]=p29−p will always be divisible by 29, as it is prime.
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