Let ‘P’ be a prime number such that P ge 11, let n = P! + 1 . The number of primes in the list n + 1, n + 2, n + 3,...., n + P - 1 is

Question

Let ‘P’ be a prime number such that P ge 11, let n = P! + 1 . The number of primes in the list n + 1, n + 2, n + 3,...., n + P - 1 is (A) P - 1 (B) 2 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

For 1 le i le P-1 ; P ! is divisible by i + 1 Then n + i = P! + (i + 1) is divisible by i + 1 for 1 le i le P - 1 ∴ None of n + 1 , n + 2 ,......n + P - 1 is a Prime

Q. Let ‘P’ be a prime number such that $P \geq 11$ ,
let $n = P! + 1$ .
The number of primes in the list $n + 1, n + 2, n + 3, ...., n + P - 1$ is

P - 1

2

1

0

For $1 \leq i \leq P - 1; P!$ is divisible by i + 1
Then n + i = P! + (i + 1) is divisible by i + 1 for $1 \leq i \leq P - 1$
$∴$ None of $n + 1, n + 2, ...... n + P - 1$ is a Prime

Q. Let ‘P’ be a prime number such that P≥11, let n=P!+1 . The number of primes in the list n+1,n+2,n+3,....,n+P−1 is

P - 1

2

1

0

For 1≤i≤P−1;P! is divisible by i + 1 Then n + i = P! + (i + 1) is divisible by i + 1 for 1≤i≤P−1 ∴ None of n+1,n+2,......n+P−1 is a Prime

Q. Let ‘P’ be a prime number such that $P \geq 11$ ,
let $n = P! + 1$ .
The number of primes in the list $n + 1, n + 2, n + 3, ...., n + P - 1$ is

For $1 \leq i \leq P - 1; P!$ is divisible by i + 1
Then n + i = P! + (i + 1) is divisible by i + 1 for $1 \leq i \leq P - 1$
$∴$ None of $n + 1, n + 2, ...... n + P - 1$ is a Prime