The International Mathematical Olympiad "Tuymaada" ...
Even students from Romania participated there. We read on Andrei Alex ECKSTEIN's blog : "The Tuymaada International Multidisciplinary Olympiad is a competition held annually in Yakutsk, Sakha Republic (Russian Federation). The competition has sections: mathematics, computer science, physics and chemistry. There are two days of competition. Participation is numerically small, for example in 2013 150 students from 6 countries participated, including Romania. Although very far away, participation in the competition is justified by the exceptional quality of the problems. Since 2000, the competition has had a section dedicated to juniors."
Follow the path :
HOME$\rightarrow$PROBLEME DIVERSE$\rightarrow$CONCURSURI$\rightarrow$"TUYMAADA"
I was interested in "INTERNATIONAL OLYMPIAD "TUYMAADA-2025" (mathematics) Second day" Problems 6 from both the Seniors and Juniors :
Senior League 6. In a sequence $(x_n)$, the number $x_1$ is positive and
rational, and
$x_{n+1} = \frac{\{nx_n\}}{n}$ for $n\geqslant 1$
($\{a\}$ denotes the fractional part of $a$). Prove that this
sequence contains only finitely many non-zero terms
and their sum is an integer.
(V. Kolezhuk, O, Tarakanov )
Junior League 6. In a sequence $(x_n)$, the first number $x_1$ is positive,
and
$x_{n+1} =\frac{\{ nx_n\}}{ n}$ for $n\geqslant 1$
($\{a\}$ denotes the fractional part of $a$). Prove that the
sequence does not contain zeroes if and only if $x_1$ is
irrational.
(V. Kolezhuk, O, Tarakanov )
$\blacklozenge$CiP Comments
We will refer to these problems by the notations S6, J6 respectively.
$\blacklozenge$Problem J6 has a logical aspect
$$\forall n\;(x_n\neq 0)\;\Leftrightarrow\; x_1\notin\mathbb{Q}$$
The statement
$x_1\notin \mathbb{Q}\;\Rightarrow\;\forall n\;(x_n\neq 0)$
is almost trivial: from $\{nx_n\}=nx_n-[nx_n]$ we have $x_{n+1}=\color{Red}{x_n}-\frac{[nx_n]}{n}$, so
$x_n\notin \mathbb{Q}\;\Rightarrow\;x_{n+1}\notin \mathbb{Q}$, hence $x_{n+1}\neq 0$. Thus we have $\forall n\;(x_n \neq 0).$
For statement
$\forall n\;(x_n \neq 0)\;\Rightarrow\;x_1\notin \mathbb{Q}$
we prove its converse instead
$x_1 \in \mathbb{Q}\;\Rightarrow\;\exists n\;(x_n=0) \tag{SJ_1}$
that is a common requirement for both problems J6, S6.
$\blacklozenge$Let's look at some examples.
Example 1 $x_1=\frac{2}{3}$
$x_2=\frac{\{x_1\}}{1}=\left\{\frac{2}{3}\right \}=\frac{2}{3}\;;\;x_3=\frac{\{2x_2\}}{2}=\frac{\left \{\frac{4}{3}\right \}}{2}=\frac{\frac{1}{3}}{2}=\frac{1}{6}\;;\;x_4=\frac{\{3x_3\}}{3}=\frac{\left \{\frac{3}{6}\right \}}{3}=\frac{\frac{3}{6}}{3}=\frac{1}{6}$
$x_5=\frac{\{4x_4\}}{4}=\frac{\left \{\frac{4}{6}\right \}}{4}=\frac{\frac{4}{6}}{4}=\frac{1}{6}\;;\;x_6=\frac{\{5x_5\}}{5}=\frac{\left \{\frac{5}{6}\right \}}{5}=\frac{\frac{5}{6}}{5}=\frac{1}{6}$
$x_7=\frac{\{6x_6\}}{6}=\frac{\{1\}}{6}=0$ and from here on out, all $x_n=0,\;n\geqslant 8$.
The sum of the nonzero terms is
$$x_1+x_2+x_3+x_4+x_5+x_6=\frac{2}{3}+\frac{2}{3}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=2 \tag{S_Ex_1}$$
We will see later the connection with Egyptian writing
$$\frac{2}{3}=\frac{1}{2}+\frac{1}{6} \tag{E_Ex_1}$$
Example 2. $x_1=\frac{3}{4}$
$x_2=\frac{\{x_1\}}{1}=\left\{\frac{3}{4}\right\}=\frac{3}{4}\;;\;x_3=\frac{\{2x_2\}}{2}=\frac{\left\{\frac{6}{4}\right \}}{2}=\frac{\frac{2}{4}}{2}=\frac{1}{4}\;;\;x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{3}{4}\right \}}{3}=\frac{\frac{3}{4}}{3}=\frac{1}{4}$
$x_5=\frac{\{4x_4\}}{4}=\frac{\{1\}}{4}=0$ and from here on out, all $x_n=0,\;n\geqslant 6$.
The sum of the nonzero terms is
$$x_1+x_2+x_3+x_4=\frac{3}{4}+\frac{3}{4}+\frac{1}{4}+\frac{1}{4}=2 \tag{S_Ex_2}$$
and the representation as a sum of Egyptian fractions of $x_1$ is
$$\frac{3}{4}=\frac{1}{2}+\frac{1}{4} \tag{E_Ex_2}$$
Example 3. $x_1=\frac{17}{5}$
$x_2=\frac{\{x_1\}}{1}=\left\{ \frac{17}{5}\right \}=\frac{2}{5}\;;\;x_3=\frac{\{2x_2\}}{2}=\frac{\left\{\frac{4}{5}\right\}}{2}=\frac{\frac{4}{5}}{2}=\frac{2}{5}\;;\;x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{6}{5}\right\}}{3}=\frac{\frac{1}{5}}{3}=\frac{1}{15}$
$x_5=\frac{\{4x_4\}}{4}=\frac{\left\{\frac{4}{15}\right\}}{4}=\frac{1}{15}\;;\;x_6=\frac{\{5x_5\}}{5}=\frac{\left\{\frac{5}{15}\right\}}{5}=\frac{\frac{5}{15}}{5}=\frac{1}{15}\;;\;x_7=\frac{\{6x_6\}}{6}=\frac{\left\{\frac{6}{15}\right\}}{6}=\frac{\frac{6}{15}}{6}=\frac{1}{15}$
$x_8=\frac{\{7x_7\}}{7}=\frac{\left\{\frac{7}{15}\right\}}{7}=\frac{\frac{7}{15}}{7}=\frac{1}{15}\;;\;x_9=\frac{\{8x_8\}}{8}=\frac{\left\{\frac{8}{15}\right\}}{8}=\frac{\frac{8}{15}}{8}=\frac{1}{15}\;;\;x_{10}=\frac{\{9x_9\}}{9}=\frac{\left\{\frac{9}{15}\right\}}{9}=\frac{\frac{9}{15}}{9}=\frac{1}{15}$
and so on... $x_{11}=\frac{1}{15}=x_{12}=x_{13}=x_{14}=x_{15}\left (=\frac{\{14x_{14}\}}{14}=\frac{\left\{\frac{14}{15}\right\}}{14}=\frac{\frac{14}{15}}{14}=\frac{1}{15}\right )$
but(!) $x_{16}=\frac{\{15x_{15}\}}{15}=\frac{\{1\}}{15}=0$ and from here on out, $x_n=0,\;n\geqslant 17$.
The sum of the nonzero terms is
$$x_1+x_2+x_3+x_4+\dots+x_{15}=\frac{17}{5}+2\cdot\frac{2}{5}+12\cdot \frac{1}{15}=5\tag{S_Ex_3}$$
and the representation as a sum of Egyptian fractions of $x_1$ is
$$\frac{17}{5}=3+\frac{1}{3}+\frac{1}{15} \tag{E_Ex_3}$$
Example 4. $x_1=\frac{7}{6}$
$x_2=\frac{\{x_1\}}{1}=\left \{\frac{7}{6}\right\}=\frac{1}{6};$ without further calculation, so in the previous examples, we have
$x_3=x_4=x_5=x_6=\frac{1}{6},\;x_7=\frac{\{6x_6\}}{6}=\frac{\{1\}}{6}=0$, and furher $x_n=0,\;n\geqslant 8$.
The sum of the nonzero term is
$$x_1+x_2+\dots+x_6=\frac{7}{6}+5\cdot \frac{1}{6}=2 \tag{S_Ex_4}$$
and $x_1$ has the representation as the sum of Egyptian fractions
$$\frac{7}{6}=1+\frac{1}{6} \tag {E_Ex_4}$$
Example 5. $x_1=\frac{3}{7}$
We quickly see that $x_3=x_2=\{x_1\}=\frac{3}{7}$; then that $x_{11}=x_{10}=\dots=x_4=\frac{\{3x_3\}}{3}=\frac{\left\{\frac{9}{7}\right\}}{3}=\frac{\frac{2}{7}}{3}=\frac{2}{21}$;
and the eye, increasingly experienced, sees that $x_{231}=x_{230}=\dots =x_{12}=\frac{\{11 x_{11}\}}{11}=\frac{\left\{\frac{22}{21}\right\}}{11}=\frac{\frac{1}{21}}{11}=\frac{1}{231}$.
We're almost done, because $x_{232}=\frac{\{231x_{231}\}}{231}=\frac{\{1\}}{231}=0$ and $x_n=0,\;n\geqslant 232$.
The sum of the nonzero terms is
$$x_1+x_2+x_3+(x_4+x_5+\dots+x_{11})+(x_{12}+x_{13}+\dots+x_{231})=3\cdot \frac{3}{7}+8\cdot \frac{2}{21}+220\cdot \frac{1}{231}=3 \tag{S_Ex_5}$$
and $x_1$ has the representation as the sum of Egyptian fractions
$$\frac{3}{7}=\frac{1}{3}+\frac{1}{11}+\frac{1}{231} \tag{E_Ex_5}$$
$\blacklozenge$fghn
<end CiP comments>$\blacklozenge$
<de continuat>