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A positive integer $n$ will be called monotonous, if its decimal digits are in ascending or descending order.



For example $1111, 1122, 113355, 998850, 66655522$ and $332$ are monotonous numbers but $121, 2017$ and $5555554444445$ are not monotonous numbers.



How can I prove that for every positive integer $n$, there is a positive integer $m$ such that $nm$ is a monotonous number?

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closed as off-topic by Henrik, Glorfindel, JonMark Perry, Antonios-Alexandros Robotis, user91500 Aug 12 '17 at 11:47

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  • 3
    $\begingroup$ Hint: there is even always a multiple of the form 11...1100...00 $\endgroup$ – Henning Makholm Aug 12 '17 at 7:33
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$\color{Purple}{\text{ The answer is }}$ $\color{Green}{\text{ YES }}$.



Lemma(Euler): Let $n$ to be a positive ineger such that $\gcd(n,10)=1$; then we have:

$$ 10^{\varphi(n)} \overset{n}{\equiv} 1 \ ,$$ where $\varphi(n)$ euler's function; i.e. :

$$ \varphi(n) = \text{Card}\Big( \{ a \in \mathbb{N} \ | \ 1 \leq a \leq n \ \ \ \text{&} \ \ \ \gcd(a,n)=1 \} \Big) \ .$$



At first assume that $\gcd(n,10)=1$;
then let $k:=\dfrac{10^{\varphi(n)}-1}{n}$, then:

$$kn = \underbrace{9 9 ... 9 9}_{\varphi(n)-\text{ times}}.$$



Now let $N$ to be a positive arbitrary integer, then there exist integers $r \in \mathbb{N} \cup \{ 0 \} $ & $s \in \mathbb{N} \cup \{ 0 \} $ and $n$; such that $\gcd(n,10)=1$ and $N=2^r 5^s n$.

Let $t:=\max \{ r,s \} $, and let:

$$k:=\dfrac{10^{\varphi(n)}-1}{n}.10^t \ ; $$

then we have:

$$kN = \underbrace{9 9 9 ... 9 9 9}_{\varphi(n)-\text{ times}} \underbrace{0 0 0 ... 0 0 0}_{t-\text{ times}} .$$

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    $\begingroup$ Note that Euler's theorem is unnecessary for proving existence. One can just use the pigeonhole principle on residues of $10^s$ mod $n$. $\endgroup$ – Erick Wong Aug 12 '17 at 12:43
  • $\begingroup$ @ Erick Wong , yes are right. $\endgroup$ – Davood Khajehpour Aug 12 '17 at 12:45

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