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Problem A. 509. (May 2010)

A. 509. Prove that there exists a real number c>0 with the following property: among arbitrary, pairwise distinct positive integers a_1,a_2,\ldots,a_n (n\ge3), there are three whose least common multiple is at least c.n2.99.

(5 pont)

Deadline expired on June 10, 2010.


Sorry, the solution is available only in Hungarian. Google translation

Megoldásvázlat. Az állítást három olyan esetben bizonyítjuk, amelyek együttesen lefedik az összes lehetséges sorozatot.

1. eset: Van olyan 1\lek\len-2, amire \frac{a_{k+2}}{a_k}<1+2000n^{-1+\frac1{300}}.


[a_k,a_{k+1},a_{k+2}] =
\frac{a_ka_{k+1}a_{k+2} \cdot (a_k,a_{k+1},a_{k+2})}{
(a_k,a_{k+1})\cdot(a_k,a_{k+2})\cdot(a_{k+1},a_{k+2})}\ge


\ge \frac{a_k^2a_{k+1}\cdot1}{(a_{k+1}-a_k)(a_{k+2}-a_k)(a_{k+2}-a_{k+1})} =


=
\frac1{\big(\frac{a_{k+1}}{a_k}-1\big)\big(\frac{a_{k+2}}{a_k}-1\big)\big(\frac{a_{k+2}}{a_{k+1}}-1\big)}
> \frac1{\big(\frac{a_{k+2}}{a_k}-1\big)^3}
> \frac1{\Big(2000n^{-1+\frac1{300}}\Big)^3}
> \frac{n^{2,99}}{10^{10}}.

2. eset: n\ge2000, és \frac{a_{k+2}}{a_k}\ge 1+2000n^{-1+1000\frac1{300}} minden 1\lek\len-2 esetén.


a_n
\ge \frac{a_3}{a_1} \cdot \frac{a_5}{a_3} \cdot\ldots\cdot
\frac{a_{2\lceil n/2\rceil-1}}{a_{2\lceil n/2\rceil-3}}
\ge \left(1+2000n^{-1+\frac1{300}}\right)^{\frac{n-1}/2}.

Mivel \frac{n-1}{1800}>1, a Bernoulli-egyenlőtlenségből


\left(1+2000n^{-1+\frac1{300}}\right)^{\frac{n-1}{1800}} \ge
1+\frac{n-1}{1800}\cdot 2000n^{-1+\frac1{300}} >
\frac{n}{2000}\cdot 2000n^{-1+\frac1{300}} >
n^{\frac1{300}}.

Tehát,


[a_1,a_2,a_n] \ge a_n > 
\left(\left(1+2000n^{-1+\frac1{300}}\right)^{\frac{n-1}{1800}}\right)^{900} \ge
\left(n^{\frac1{300}}\right)^{900} = n^3.

3. eset: n<2000.


[a_1,a_2,a_3] > 1 > \frac{n^3}{10^{10}}.

Az állítás tehát minden sorozatra teljesül a c=10-10 választással.


Statistics:

5 students sent a solution.
5 points:Backhausz Tibor, Bodor Bertalan, Éles András, Nagy 235 János, Nagy 648 Donát.

Problems in Mathematics of KöMaL, May 2010