1969 AHSME Problems/Problem 23

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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. $\fbox{A}$

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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