Given that p is a prime number- p2p27-1-p is always divisible by-

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Quantitative Aptitude

Fermat Theorem

Given that p ...

Question

Given that p is a prime number, $p^{2} (p^{27} - \frac{1}{p})$ is always divisible by?

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Solution

The correct option is D none of these
According to Fermat's theorem "If M is a prime number, then for any integer N, $N^{M} - N$ is divisible by M."

Open the brackets, $p^{2} (p^{27} - \frac{1}{p}) = p^{29} - p$ will always be divisible by 29, as it is prime.

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Given that p is a prime number, p2(p27−1p) is always divisible by?

The correct option is D none of theseAccording to Fermat's theorem "If M is a prime number, then for any integer N, NM−N is divisible by M." Open the brackets, p2(p27−1p)=p29−p will always be divisible by 29, as it is prime.

Given that p is a prime number, $p^{2} (p^{27} - \frac{1}{p})$ is always divisible by?

The correct option is D none of these
According to Fermat's theorem "If M is a prime number, then for any integer N, $N^{M} - N$ is divisible by M."

Open the brackets, $p^{2} (p^{27} - \frac{1}{p}) = p^{29} - p$ will always be divisible by 29, as it is prime.