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Standard XII

Mathematics

Summation by Sigma Method

Let p be a pr...

Question

Let p be a prime number such that $p > 11 .$ Let $n = p! + 1 .$ The number of prime numbers in the list $n + 1, n + 2, n + 3, \dots$ $n + p - 1$ is/ are

p - 1

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Solution

The correct option is D $0$
$n = p! + 1$

$n + 1 = p! + 1 + 1 = p! + 22 | p! \Rightarrow 2 | n + 1$

$n + 2 = p! + 3 \Rightarrow 3 | p! S o 3 | n + 2$

$n + 3 = p! + 44 | p! S o 4 | n + 3$

$n + p - 1 = p! + p p | p! S o p | p! + p$

Hence the are no prime numbers in the given list

$D, 0$

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Let p be a prime number such that p>11. Let n=p!+1. The number of prime numbers in the list n+1,n+2,n+3,… n+p−1 is/ are

The correct option is D 0n=p!+1 n+1=p!+1+1=p!+22|p!⇒2|n+1 n+2=p!+3⇒3|p! So 3|n+2 n+3=p!+44|p! So 4|n+3 n+p−1=p!+pp|p! So p|p!+p Hence the are no prime numbers in the given list D,0

Let p be a prime number such that $p > 11 .$ Let $n = p! + 1 .$ The number of prime numbers in the list $n + 1, n + 2, n + 3, \dots$ $n + p - 1$ is/ are

The correct option is D $0$
$n = p! + 1$

$n + 1 = p! + 1 + 1 = p! + 22 | p! \Rightarrow 2 | n + 1$

$n + 2 = p! + 3 \Rightarrow 3 | p! S o 3 | n + 2$

$n + 3 = p! + 44 | p! S o 4 | n + 3$

$n + p - 1 = p! + p p | p! S o p | p! + p$

Hence the are no prime numbers in the given list

$D, 0$