Prove that there exists a real number $C > 1$ with the following property. Whenever $n > 1$ and $a_0 < a_1 < a_2 <\cdots < a_n$ are positive integers such that $\frac{1}{a_0},\frac{1}{a_1} \cdots \frac{1}{a_n}$ form an arithmetic progression, then $a_0 > C^n$.
Problem
Source: CIIM 2015, India Postal Coaching 2015
Tags: number theory, bounded
04.12.2015 16:18
Put $lcm(a_0,a_1,\dots,a_n)=q$ and $\frac1{a_i}=\frac{r_i}{q}$, then $r_0>r_1>\dots>r_n$ is an arithmetic sequence and $q/r_i=a_i$ $\implies$ $r_i|q$ for each $i$. So we have $a_0=\frac{q}{r_0}$ where $lcm(r_0,r_1,\dots,r_n)|q$. So $a_0\ge \frac{lcm(r_0,r_1,\dots,r_n)}{r_0}$. Now we wish to estimate $\frac{lcm(r_0,r_1,\dots,r_n)}{r_0}$. This follows from the following lemma (easy to prove using reasoning similar to Legendre's formula): if $0<u_1<u_2<\dots$ is an arithmetic progression of integers relatively prime to its difference, and $L=lcm(u_1,u_2,\dots,u_n)$, $T=\frac{u_1u_2\dots u_n}{n!}$, then for any prime $p$, $v_p(T)\le v_p(L)$, and as a corollary, $T\le L$.
Claim: for any arithmetic sequence $r_0,r_1,\dots,r_n>0$, we have $\frac{lcm(r_0,r_1,\dots,r_n)}{r_0}\ge \frac{r_1r_2r_3\dots r_{[n/2]}}{([n/2]+1)!}$.
From here, we are done, because $\frac{r_1r_2r_3\dots r_{[n/2]}}{([n/2]+1)!}\ge \binom{n}{[n/2]}>\frac{2^n}{n+1}$
04.12.2015 16:24
btw here is the official solution from KOMAL http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A652&l=en
04.12.2015 16:35
can anyone please give me the details of this competition?
04.12.2015 16:37
i_atmmaths wrote: can anyone please give me the details of this competition? Which competiotion? Do you mean KOMAL?
04.12.2015 16:52
Satyaprakash2009rta wrote: i_atmmaths wrote: can anyone please give me the details of this competition? Which competiotion? Do you mean KOMAL? yes
04.12.2015 16:56
You can visit to its official website. http://www.komal.hu