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Difference between revisions of "1989 IMO Problems/Problem 5"

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==Problem==
 
==Problem==
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Let <math>n\geq3</math> and consider a set <math>E</math> of 2n−1 distinct points on a circle. Suppose that exactly <math>k</math> of these points are to be colored black. Such a coloring is “good” if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly <math>n</math> points from set <math>E</math>. Find the smallest value of <math>k</math> such that every such coloring of <math>k</math> points of <math>E</math> is good.
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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.
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==Solution==
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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>.
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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).
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Edit: to elementarize the <math>\lim_{n \to \infty} \frac{\pi(n)}{n} = 0</math> part:
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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>.
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This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [https://aops.com/community/p372297]
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== See Also == {{IMO box|year=1989|num-b=4|num-a=6}}
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[[Category:Olympiad Number Theory Problems]]

Revision as of 11:03, 30 January 2021

Problem

Prove that for each positive integer $n$ there exist $n$ consecutive positive integers none of which is an integral power of a prime number.

Solution

There are at most $1+\sqrt[2]{n}+\sqrt[3]{n}+\sqrt[4]{n}+...+\sqrt[\left\lfloor \log_2(n)\right\rfloor]{n} \leq 1+ \sqrt n log_2(n)$ 'true' powers $m^k , k\geq 2$ in the set $\{1,2,...,n\}$. So when $p(n)$ gives the amount of 'true' powers $\leq n$ we get that $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.

Since also $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$, we get that $\lim_{n \to \infty} \frac{p(n)+\pi(n)}{n} = 0$. Now assume that there is no 'gap' of lenght at least $k$ into the set of 'true' powers and the primes. Then this would give that $\frac{p(n)+\pi(n)}{n} \geq \frac{1}{k}$ for all $n$ in contrary to the above (at east this proves a bit more).

Edit: to elementarize the $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$ part: Look $\mod (k+1)!$. Then all numbers in the residue classes $2,3,4,...,k+1$ are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length $k$, there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each $(k+1)!$ numbers, again contradicting $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.

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
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