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A stone placed on the $X-Y$ plane moving along points with positive integer co-ordinates. When the stone is at $(x,y)$, it can be moved to $(3x,y)$ or $(x,3y)$ or $(x-z,y-z)$ where $z \in \mathbb{Z}^+$. Find all possible value $m,n$ of order pair $(m,n)$, where $m, n \in \mathbb{Z}^+$ such that the stone can be moved from $(m,n)$ to $(0, 0)$.

Is my answer correct ?

Consider 2 cases :

1) $x\not= y$, WLOG, $y>x$.

Let $l \in \mathbb{N}_0$ such that $3^lx \leq y<3^{l+1}x $

Since $3x-x$ is even, $x-y=(x-z)-(y-z)$ and $0-0=0$, so $m,n$ have the same parity.

$\exists t \in \mathbb{N}_0$, $y=3^lx+2t$

Since $y<3^{l+1}x $ so $t<3^lx$.

Thus, $(x,y) = (x,3^lx+2t) \rightarrow (3^lx,3^lx+2t)$

As $t<3^lx$, so $(3^lx,3^lx+2t) \rightarrow (t,3t) \rightarrow (3t,3t) \rightarrow (0, 0)$

2) $x=y$ we get $(x,y)\rightarrow (x-y,y-y)\rightarrow (0, 0)$

Answer : All positive integers, $m, n$ that have the same parity or $m\equiv n(\bmod{2})$.

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  • $\begingroup$ Sounds like we are allowed to make a sequence moves as long as each move obeys the rules? Are we to assume that $x\ge0,y\ge0$ throughout, or does the playing arena also contain points with negative coordinates (not that it matters much because once the stone dips into the negative side it can never come back?). $\endgroup$ – Jyrki Lahtonen Aug 24 '17 at 13:17
  • $\begingroup$ Anyway, your method looks fine to me. $\endgroup$ – Jyrki Lahtonen Aug 24 '17 at 13:19
  • $\begingroup$ take (x,y)=(1,3), we then get (3,3),(1,9), and and pairing with a difference of two so it doesn't go directly to (0,0).I think this was @JyrkiLahtonen's point. $\endgroup$ – user451844 Aug 24 '17 at 13:23
  • $\begingroup$ @ Jyrki Lahtonen♦, Edited, thank you. $\endgroup$ – carat Aug 24 '17 at 13:31
  • $\begingroup$ Ok. I was slightly worried about starting positions like $(x,y)=(2,0)$ that satisfies the parity condition, but won't let you move away from the $x$-axis. $\endgroup$ – Jyrki Lahtonen Aug 24 '17 at 13:32
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You're right, but there is an easier way of going about it (in my opinion).

You're totally right to say that $m$ and $n$ must have the same parity, since multiplying an integer by $3$ preserves parity, and reducing $m$ and $n$ by the same integer $z$ preserves their relative parity. Ergo, since $0$ and $0$ have the same parity, so too must $m$ and $n$.

There are then, as you say, two cases. The case $m=n$ is trivial, and in the case $m\neq n$ (setting $m<n$ w.l.o.g.), one can always reduce $(m,n)$ to $(1,n-m+1)$, where $n-m+1$ is odd because we insist $m\equiv n\ (\mathrm{mod\ 2})$. Then we can just repeat the following two steps:

  • Triple the $1$: $(1,x) \to (3,x)$
  • Subtract $2$ from each integer: $(3,x) \to (1,x-2)$

Repeat until we are left with $(1,3)$, which we must reach eventually. This then goes to $(3,3)$ and hence we reach $(0,0)$.

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  • $\begingroup$ Nice and easy. Thank you very much ! $\endgroup$ – carat Aug 24 '17 at 13:42

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