Difference between revisions of "1989 IMO Problems/Problem 5"
(→Problem) |
|||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | + | Prove that for each positive integer <math> n</math> there exist <math> n</math> consecutive positive integers none of which is an integral power of a prime number. | |
+ | |||
+ | ==Solution== | ||
+ | There are at most <math>1+\sqrt[2]{n}+\sqrt[3]{n}+\sqrt[4]{n}+...+\sqrt[\left\lfloor \log_2(n)\right\rfloor]{n} \leq 1+ \sqrt n log_2(n)</math> 'true' powers <math>m^k , k\geq 2</math> in the set <math>\{1,2,...,n\}</math>. So when <math>p(n)</math> gives the amount of 'true' powers <math>\leq n</math> we get that <math>\lim_{n \to \infty} \frac{p(n)}{n} = 0</math>. | ||
+ | |||
+ | Since also <math>\lim_{n \to \infty} \frac{\pi(n)}{n} = 0</math>, we get that <math>\lim_{n \to \infty} \frac{p(n)+\pi(n)}{n} = 0</math>. Now assume that there is no 'gap' of lenght at least <math>k</math> into the set of 'true' powers and the primes. Then this would give that <math>\frac{p(n)+\pi(n)}{n} \geq \frac{1}{k}</math> for all <math>n</math> in contrary to the above (at east this proves a bit more). | ||
+ | |||
+ | Edit: to elementarize the <math>\lim_{n \to \infty} \frac{\pi(n)}{n} = 0</math> part: | ||
+ | Look <math>\mod (k+1)!</math>. Then all numbers in the residue classes <math>2,3,4,...,k+1</math> are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length <math>k</math>, there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each <math>(k+1)!</math> numbers, again contradicting <math>\lim_{n \to \infty} \frac{p(n)}{n} = 0</math>. | ||
+ | |||
+ | This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [https://aops.com/community/p372297] | ||
+ | |||
+ | == See Also == {{IMO box|year=1989|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 12:03, 30 January 2021
Problem
Prove that for each positive integer there exist consecutive positive integers none of which is an integral power of a prime number.
Solution
There are at most 'true' powers in the set . So when gives the amount of 'true' powers we get that .
Since also , we get that . Now assume that there is no 'gap' of lenght at least into the set of 'true' powers and the primes. Then this would give that for all in contrary to the above (at east this proves a bit more).
Edit: to elementarize the part: Look . Then all numbers in the residue classes are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length , there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each numbers, again contradicting .
This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]
See Also
1989 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |