Let p be a prime number such that p - 3- Let x - p- - 1- The number of primes in the list n -1- n - 2- n - 3- n - p - 1 is

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Standard IX

Mathematics

Distributive Property

Let p be a pr...

Question

Let $p$ be a prime number such that $p \geq 3$ . Let $x = p! + 1$ . The number of primes in the list $n + 1, n + 2, n + 3, . . . ., n + p - 1$ is

$p - 1$

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$2$

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$1$

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$0$

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Solution

The correct option is D
$0$

Explanation for the correct option:
Finding the number of primes in the list $n + 1, n + 2, n + 3, . . . ., n + p - 1$ is
Given: $p$ be a prime number such that $p \geq 3$
Also: $x = p! + 1$
For
$1 \leq k \leq p - 1 x + k = p! + k + 1 2 \leq k + 1 \leq p ∵ 1 \leq k \leq p - 1 \Rightarrow x + k = (1, 2, 3, . . . . ., k + 1, . . . p) + (k + 1)$
Since there are no prime number in the above equation
Hence the correct option is (D)

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Let p be a prime number such that p≥3. Let x=p!+1. The number of primes in the list n+1,n+2,n+3,....,n+p–1 is

Let $p$ be a prime number such that $p \geq 3$ . Let $x = p! + 1$ . The number of primes in the list $n + 1, n + 2, n + 3, . . . ., n + p - 1$ is