Let $[a]$ represent the largest integer less than or equal to $a$, for any real number $a$. Let $\{a\} = a - [a]$. Are there positive integers $m,n$ and $n+1$ real numbers $x_0,x_1,\hdots,x_n$ such that $x_0=428$, $x_n=1928$, $\frac{x_{k+1}}{10} = \left[\frac{x_k}{10}\right] + m + \left\{\frac{x_k}{5}\right\}$ holds? Justify your answer.
Problem
Source: 2019 CSMO Grade 11 Problem 1
Tags: algebra
WypHxr
30.07.2019 14:44
By some calculation, we find $x_{4i}=428+40i$, $x_{4i+1}=426+10(4i+1)$, $x_{4i+2}=422+10(4i+2)$, $x_{4i+3}=424+10(4i+3)$, if there exists $k$, $40k=1928-428$, $4 \mid 150$, a contradiction.
FUSK
30.07.2019 15:44
Is there a rigorous proof?
IndoMathXdZ
04.08.2019 17:49
Suppose such $n$ exists. Bash the first few values of $x_i$. $x_1 = 10 \left( 42 + m + \frac{3}{5} \right) = 426 + 10m$. $x_2 = 10 \left( 42 + m + m + \frac{1}{5} \right) = 422 + 20m $. $x_3 = 10 \left( 42 + 2m + m + \frac{2}{5} \right) = 424 + 30m$. $x_4 = 10 \left( 42 + 3m + m + \frac{4}{5} \right) = 428 + 40m$. Notice that $x_1 \equiv 6 \ (\text{mod} 10) \ $, $x_2 \equiv 2 \ (\text{mod} 10 ) \ $, $x_3 \equiv 4 \ (\text{mod} 10) \ $, $x_4 \equiv 8 (\text{mod} 10)$. By induction one can prove, that this modulo behavior periodic for indices modulo $4$. So, we must have $x_{4k} = 1928$. But, we have $x_{4k} = 428 + 40mk = 1928$, but clearly $1500$ is not divisible by $40$.