1969 AHSME Problems/Problem 23

Problem

For any integer $n>1$ , the number of prime numbers greater than $n!+1$ and less than $n!+n$ is:

$\text{(A) } 0\quad\qquad \text{(B) } 1\quad\\ \text{(C) } \frac{n}{2} \text{ for n even, } \frac{n+1}{2} \text{ for n odd}\quad\\ \text{(D) } n-1\quad \text{(E) } n$

Solution

Observe that for all $k \in 1< k< n$ , since $k$ divides $n!$ , $k$ also divides $n!+k$ . Therefore, all numbers $a$ in the range $n!+1<a<n!+n$ are composite. Therefore there are 0 primes in that range. $\fbox{A}$

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1969 AHSME Problems/Problem 23

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