A calculator has only 2 buttons. The first multiplies the current value by 2, the second divides it by 3 without a remainder (so 8 becomes 2). Can you use this calculator to reach every positive integer when starting with 1?

Here is a similar question: Two button calculator

  • $\begingroup$ The question is if this is a kind of reversed Collatz conjecture or a significantly different thing. In the first case there is no proof yet. $\endgroup$
    – balazs.com
    Commented Nov 16, 2019 at 17:04

2 Answers 2


Sorry that I ignored that comment of yours in the other post.

In fact the answer is "yes": every number can be made.

Proof: Note that the number $\alpha = \log_32$ is irrational. This implies that the sequence $(\{ k\alpha\})_{k \geq 0}$ is dense in the interval $[0,1)$, where $\{\cdot\}$ denotes the fractional part.

Now let $n$ be an integer and consider the interval $[\log_3n, \log_3(n + 1))$. Its image under the map $\{ \cdot \}:\mathbb R \rightarrow [0, 1)$ contains a non-empty open set. Hence there exists infinitely many integers $k\geq0$ such that $\{ k\alpha \}$ lies in the image of that interval.

We take such a $k$ that is sufficiently large, so that $k\alpha \geq \log_3n$. Then there is an integer $m\geq0$ such that $k\alpha - m$ lies in the interval $[\log_3n,\log_3(n + 1))$.

Therefore we get the number $n$ via $\lfloor \frac{2^k}{3^m}\rfloor$.

  • $\begingroup$ Ah cool! I am guessing this doesn't give the minimal number of steps though? $\endgroup$ Commented Nov 16, 2019 at 19:51
  • 1
    $\begingroup$ @DmitryKamenetsky This is far from the minimal number of steps. For example, this algorithm needs $13$ steps to get $6$. And to get $30$ you need $98$ steps... $\endgroup$
    – WhatsUp
    Commented Nov 16, 2019 at 21:46
  • 1
    $\begingroup$ It is actually quite easy to apply this. Start with the value $1$. If the current value is less than the goal value $n$ then double it, and if it is $n+1$ or greater then divide by $3$ (without rounding down). Repeat until you get a value between $n$ and $n+1$. Let $k$ be the number of doublings you needed, and $m$ the number of divisions by $3$. This means that you can get to $n$ by doing $k$ doublings followed by $m$ divisions by $3$ with rounding down. $\endgroup$ Commented Nov 16, 2019 at 22:04

I started making the numbers progressively by hand. The even numbers are easy to compute. (Twice of another lower number). Odd numbers which are divisible by 3 also make unbroken chains (3>6>12>24>48>96 and 9>18>36>72).

  • $1$
  • 1>$2$
  • 1>2>4>8>16>5>10>$3$
  • 1>2>$4$
  • 1>2>4>8>16>$5$
  • 3>$6$
  • 4>8>16>32>64>21>$7$
  • 4>$8$
  • 7>14>28>$9$
  • 5>$10$
  • 10>20>40>13>26>52>17>34>$11$
  • 6>$12$
  • 10>20>40>$13$
  • 7>$14$
  • ??????
  • 8>$16$
  • 13>26>52>$17$
  • 9>$18$
  • 11>22>44>88>29>58>$19$
  • 10>$20$

Here is how each number upto 20 can be written. Note that I could not find a representation for 15.

  • 1
    $\begingroup$ 1>>16>5>>160>53>106>35>70>23>46>15 $\endgroup$ Commented Nov 16, 2019 at 22:22
  • $\begingroup$ @DanielMathias Nice. I wasn't going deep enough to get it. $\endgroup$ Commented Nov 16, 2019 at 22:45

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