Difference between revisions of "1969 AHSME Problems/Problem 23"

Latest revision as of 17:45, 14 November 2017

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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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 \text{(E) } n</math>
-== Answer ==
+== Solution ==
-<math>\fbox{A}</math>
+Observe that for all <math>k \in 1< k< n</math>, since <math>k</math> divides <math>n!</math>, <math>k</math> also divides <math>n!+k</math>. Therefore, all numbers <math>a</math> in the range <math>n!+1<a<n!+n</math> are composite. Therefore there are 0 primes in that range. <math>\fbox{A}</math>
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Difference between revisions of "1969 AHSME Problems/Problem 23"

Latest revision as of 17:45, 14 November 2017

Problem

Solution

See also