$2020$ positive integers are written in one line. Each of them starting with the third is divisible by previous and by the sum of two previous numbers. What is the smallest value the last number can take? A. Gribalko
Problem
Source: Tournament of Towns 2020 oral p1 (15 March 2020)
Tags: Sequence, number theory, Divisibility
Boboska
30.05.2020 15:54
Let we call this numbers with F1, F2, ......, F2020. So we have (F1+F2)|F3, ........, (F2018+F2019)|F2020. F2020=k*(F2018+F2019) For being the smallest we can say that k=1 and therefore F2019=F2018+F2017, F2018=F2017+F2016, ...... , F3=F2+F1; So this numbers will be the Fibonachi numbers (Fn=Fn-1+Fn-2). And by the Formula of it we can find F2020=((((1+sqrt(5))/2)^2020)-(((1-sqrt(5))/2)^2020))/sqrt(5) this is integer as well(because all Fibonachi numbers are integer). P.s: Sorry for not understanding writings. This is my first writing in AoPS.
parmenides51
30.05.2020 16:26
Boboska wrote: Let we call this numbers with $F_1, F_2, ......, F_{2020}$. So we have $(F_1+F_2)|F_3, ..., (F_{2018}+F_{2019})|F_{2020}$. $F_{2020}=k (F_{2018}+F_{2019})$ For being the smallest we can say that $k=1$ and therefore $F_{2019}=F_{2018}+F_{2017}, F_{2018}=F_{2017}+F_{2016}, ... , F_3=F_2+F_1$ So this numbers will be the Fibonachi numbers ($F_n=F_{n-1}+F_{n-2}$). And by the Formula of it we can find $F_{2020}=\frac{(\frac{1+\sqrt5}{2})^{2020}}{\sqrt5}-\frac{(\frac{1-\sqrt5}{2})^{2020}}{\sqrt5}$ this is integer as well (because all Fibonachi numbers are integer). P.s: Sorry for not understanding writings. This is my first writing in AoPS. fixed the latex for you, pressing quote you might see how I wrote it in latex
placing one dollar before the math and one dollar after the math
fraction a/b -> \frac{a}{b} , power a^b -> a^{b}, index a_b -> a_{b} , square root of a - > \sqrt{a}
here you might try writing with latex, while you write it
a latex tutorial inside aops is here
Ankoganit
30.05.2020 16:53
Let the numbers in $a_0,a_1,\cdots ,a_{2019}$. Since $a_1|a_2,a_2|a_3$ and so on, one may assume $a_{i}=k_ia_{i-1}$ for all $i\ge 2$. Now $a_{i-1}+a_i|a_{i+1}\implies a_{i-1}+k_ia_{i-1}|k_{i+1}k_ia_{i-1}$, so $1+k_i|k_ik_{i+1}$, and since $(1+k_i,k_i)=1$, we have $1+k_i|k_{i+1}$, and so $k_{i+1}\ge 1+k_i$. Now $a_0+a_1|a_2$ implies $a_2>a_1$, giving $k_2\ge2$, so by induction, $k_{m}\ge m$ for all $m\ge 2$. Now we note that \begin{align*}a_{2019}&=k_{2019}a_{2018}=k_{2019}k_{2018}a_{2017}=\cdots=k_{2019}\cdots k_2a_1\\
&\ge 2019\cdot 2018\cdots 2 a_1=2019!a_1\ge 2019!
\end{align*}This bound can be achieved, for example, by taking $a_i=i!$ for $0\le i\le 2019$.