Question

Given that p is a prime number. $p^2[p^{27}-\frac{1}{p}]$ is always divisible by?

• A. 25
• B. 23
• C. 27
• D. none of these

Answer 1

The correct option is D none of these

According to fermat's theorem " If P is a prime number and N is prime to P, then $N^p-N$ is divisible by P."

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