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Explore the Back-StoryThis 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

(A) $0$;

(B) $1$;

(C) $2$;

(D) more than $2$;

**Discussion:**

$51!$ 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.

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

(A) $0$;

(B) $1$;

(C) $2$;

(D) more than $2$;

**Discussion:**

$51!$ 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.

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