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I'll appreciate solution hints for the following question from a past exam in Discrete Mathematics that was administered in my university.

Denote by $\mathbb{N}$ the set of positive integers. For every $n\in\mathbb{N}$ denote by $n\%10$ the remainder of the division of $n$ by $10$, i.e. $n%10$ is the unique number satisfying: $$ \begin{align} n\%10&\in\{0,1,2,\dots,9\},\\ n-(n\%10)&=0\mod10. \end{align} $$

We postulate that the infinite sequence $n = (n_1, n_2, \dots)$ satisfies: $$ \begin{align} n_1 &= 1,\\ n_2 &= 9,\\ n_3 &= 9,\\ n_4 &= 3, \end{align} $$ and for every $i\in\mathbb{N}_1$ $n_{i+4}$ is the unit digit in the decimal representation of the sum of the preceding $4$ components of $n$: $n_{i+4}=(n_i+n_{i+1}+n_{i+2}+n_{i+3})\%10$.

Is there some $i_0\in\mathbb{N}$ such that the following four equations hold? $$ \begin{align} n_{i_0}&=7,\\ n_{i_0+1}&=3,\\ n_{i_0+2}&=6,\\ n_{i_0+3}&=7. \end{align} $$

My attempts to solve the question

I computed $n_1$ to $n_{45}$: $$ \begin{align} n':=(&1,9,9,3,2,3,7,5,7,\\ &2,1,5,5,3,4,7,9,3,\\ &3,2,7,5,7,1,0,3,1,\\ &5,9,8,3,5,5,1,4,5,\\ &5,5,9,4,3,1,7,5,6), \end{align} $$ and I extended the sequence $(7,3,6,7)$ "backwards" to a total of 27 components: $$ \begin{align} m:=(&4,3,9,5,1,8,3,7,9,\\ &7,6,9,1,3,9,2,5,9,\\ &5,1,0,5,1,7,3,6,7). \end{align} $$

I was hoping this would enable me to solve the question by one of the following ways.

  1. If it turned out that for some $i\in\{1,2,\dots,42\}$ $n'_i=m_1=4$, $n'_{i+1}=m_2=3$, $n'_{i+2}=m_3=9$, $n'_{i+3}=m_4=5$, then the answer to the original question would be: yes. However, this is not the case.

  2. If I could identify a repeating pattern in $n'$ which didn't manifest in $m$, or vice versa, then the answer to the original question might be: no. However, I was unable to identify such patterns.

  3. If I could show that one of the digits $0, 1, \dots, 9$ didn't occur in $n'$, but occurred in $m$, or vice versa, then the answer to the original question might be: no. However, all the digits occur in both $n'$ and $m$.

  4. If I were able to prove that every possible quadruple $(a,b,c,d)\in\{0,1,2,\dots,9\}^4$ eventually occurs in either $(n_1,n_2,n_3,\dots)$ or $(\dots,m_{25},m_{26},m_{27})$ (the infinite extension of $m$ "backwards"), then the answer to the original question would be: yes. However, I wasn't able to prove this.

Evan Aad
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    No need to reinvent the wheel (permutation cycle). You can (uniquely) iterate backwards (since shifting is invertible mod $10$) so the orbits are cycles, as explained in the link. That $7,3,6,7$ lies in the initial cycle follows quickly by extending it forward, as mentioned below. See here for many examples of this sort. Once you know the simple idea it is easy to solve them all. – Bill Dubuque Jul 21 '23 at 17:42
  • In case you're curious; the smallest positive integer $i$ such that $(n_i,n_{i+1},n_{i+2},n_{i+3})=(7,6,3,7)$ is $i=1553$. – Servaes Jul 22 '23 at 14:56

1 Answers1

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Hint: Try searching forward from $7,3,6,7$. (You don't have to go very far at all before something interesting happens.)

paw88789
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  • Thanks. I never thought of extending $7,3,6,7$ forward. I've extended it forward, and discovered the beginning of the first sequence: $1,9,9,3$. This implies that if $7,3,6,7,$ occurs in the sequence beginning with $1,9,9,3$, then the latter is a cyclic sequence. Now what? – Evan Aad Jul 21 '23 at 17:14
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    a) Try to prove that your sequence must be cyclic (not using the hint). b) Think about how if you know $n_k, n_{k+1}, n_{k+2}, n_{k+3}$ you can calculate $n_{k-1}$ – Ingix Jul 21 '23 at 17:21
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    Backing up is unique. – paw88789 Jul 21 '23 at 17:36