Difference between revisions of "1989 IMO Problems/Problem 5"
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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. | 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== | + | ==Solution 1== |
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>. | 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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This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [https://aops.com/community/p372297] | This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [https://aops.com/community/p372297] | ||
+ | |||
+ | ==Solution 2== | ||
+ | By Chinese Remainder theorem, there exists <math>x</math> such that: | ||
+ | <math>x \equiv -1\;mod\;p_1 q_1\ | ||
+ | x \equiv -2\;mod\;p_2 q_2\ | ||
+ | x \equiv -3\;mod\;p_3 q_3\ | ||
+ | ...\newline | ||
+ | x \equiv -n\;mod\;p_n q_n</math> | ||
+ | Where <math>p_1, p_2, ..., p_n, q_1, q_2, ..., q_n</math> are distinct primes. | ||
+ | The n consecutive numbers <math>x+1, x+2, ..., x+n</math> each have at least two prime factors, so none of them can be expressed as an integral power of a prime. | ||
== See Also == {{IMO box|year=1989|num-b=4|num-a=6}} | == See Also == {{IMO box|year=1989|num-b=4|num-a=6}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Revision as of 20:58, 4 July 2021
Contents
[hide]Problem
Prove that for each positive integer there exist consecutive positive integers none of which is an integral power of a prime number.
Solution 1
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]
Solution 2
By Chinese Remainder theorem, there exists such that: Where are distinct primes. The n consecutive numbers each have at least two prime factors, so none of them can be expressed as an integral power of a prime.
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 |