Difference between revisions of "1969 AHSME Problems/Problem 23"
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== Solution == | == 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> |
== See also == | == See also == | ||
Latest revision as of 17:45, 14 November 2017
Problem
For any integer 
, the number of prime numbers greater than 
and less than 
is:
Solution
Observe that for all 
, since 
divides 
, also divides 
. Therefore, all numbers 
in the range 
are composite. Therefore there are 0 primes in that range. 
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
