This is an objective problem 151 from TOMATO based on Consecutive composites, useful for Indian Statistical Institute Entrance Exam.
Let $n = 51! + 1$. Then the number of primes among $n+1, n+2, ... , n+50$ is
(D) more than $2$;
$51!$ is divisible by $2, 3,... 51$.
Hence $51! +2$ is divisible by $2, ... , 51! + k$ is divisible by k if $k \le 51 $
Therefore all of these numbers are composite (none of them are primes).
Answer is (A)
Note: The above problem can be further extended to say that for any natural number of n, we can have n consecutive composite numbers.