Question

Let P be a prime number such that \[p \ge 11\]. Let n = p!+1.Find the number of primes in the list n+1,n+2,n+3,...n+P−1.
A. p -1
B. 2
C. 1
D. None of these

Answer 1

Hint: First we will write an inequality that represents all numbers less than p and greater than or equal to 1 using a variable. Using the given equation, we will check whether p! is divisible by n + i, where i is an integer. Then we compute all prime numbers.

Complete step by step solution: Assume a positive integer i such that it is less than p and greater than or equal to 1.
In other words, i is less than or equal to p - 1 and greater than or equal to 1.

The mathematical representation is
\[1 \le i \le p - 1\]
Now adding 1 on each side of the inequality:
\[ \Rightarrow 2 \le i + 1 \le p\]

Given equation is
n = p!+1
Adding i on both sides of the equation:
n + i = p!+1 + i …..(i)
We know i is an integer such that \[2 \le i + 1 \le p\].
Thus \[p! = 1 \cdot 2 \cdot 3 \cdot \cdots \left( {1 + i} \right) \cdots p\].
This implies p! is divisible by \[\left( {1 + i} \right)\].
Taking common \[\left( {1 + i} \right)\] from the right side expression of (i)
\[n{\rm{ }} + {\rm{ }}i{\rm{ }} = {\rm{ }}\left( {1 + i} \right)\left( {1 \cdot 2 \cdot 3 \cdot \cdots i \cdot \left( {i + 2} \right) \cdots p + 1{\rm{ }}\;} \right)\]
Therefore \[p! + 1 + i\] is divisible by \[\left( {1 + i} \right)\].
Since n + i = p!+1 + i , so n + i is also divisible by \[\left( {1 + i} \right)\] where i is a positive integer.
Putting i = 1, 2, 3,…, p-1 in n + i
n + 1 , n + 2, n + 3, …., n + p -1 are divisible by 1+1, 1+2, 1+ 3, ….., 1 + p – 1 respectively.

n + 1 , n + 2, n + 3, …., n + p -1 are not a prime number.

Option ‘D’ is correct

Note: If x is a positive integer less than y, then y! is divisible by x. Students often mistake the concept of divisibility of p!. They thought p! is divisible by p. But p! is divisible by all positive integers less than or equal to p.