This puzzle deals with positive integers in decimal representation. From every integer you can move to one or two or three other integers. The allowed moves for integer $n\ge1$ are as follows:

  • You may double the number (that is, $n$ becomes $2n$).
  • If the rightmost digit in the decimal representation is $4$, you may remove this rightmost digit.
  • If the rightmost digit in the decimal representation is $0$, you may remove this rightmost digit.

For instance, starting with the integer $n=227$ you could make the following moves: $$227\to454\to45\to90\to9\to18\to36\to72\to144\to14\to1\to2\to4$$

Your task: Show that starting with an arbitrary integer $n\ge1$, you can eventually reach the cosmic goal integer $4$ (by repeatedly applying these three moves).

(The title of this puzzle was chosen as repercussion of the "Four is Cosmic!" puzzle.)

  • $\begingroup$ I'm pretty sure the same is true for 2 or 8 as well. $\endgroup$ Mar 20, 2016 at 17:58
  • $\begingroup$ My suggestion: prove that you can get from n to n-1 in a finite number of moves, use induction down to 4, then provide special cases for 1, 2, and 3 (1 and 2 are trivial, 3 is 3, 6, 12, 24, 2, 4) $\endgroup$
    – hobbs
    Mar 20, 2016 at 19:57

2 Answers 2


Let the starting number be $n$. Consider the case where $n < 10$.

$$ \begin{align} 1 &\to 2 \to 4 \\ 2 &\to 4 \\ 3 &\to 6 \to 12 \to 24 \to 2 \to 4\\ 4 \\ 5 &\to 10 \to 1 \to 2 \to 4 \\ 6 &\to 12 \to 24 \to 2 \to 4\\ 7 &\to 14 \to 1 \to 2 \to 4\\ 8 &\to 16 \to 32 \to 64 \to 6 \to 12 \to 24 \to 2 \to 4\\ 9 &\to 18 \to 36 \to 72 \to 144 \to 14 \to 1 \to 2 \to 4 \end{align} $$

Now consider general $n > 0$, not ending with the digit $0$ (remove all trailing zeros before beginning).

Let a step from $n$ be defined as the sequence to get from $n$ to a number $m$ for which exactly one trailing digit is removed. For convenience, we write this functionally as $s(n) = m$. The next step is a step from $m$, and we can extend this to a sequence of steps.

Now, let $n = 10k + d$, where $k$ is an integer and $0 \leq d < 10$.

If $d = 0$, we remove the trailing 0 to get $s(n) = \frac{n}{10}$ .

If $d \in \{1,2,3,6,8\}$, first consider $d=1$. Since $s(n) = \frac{(10k+1)4-4}{10} = 4k$, $s(n)$ is even. Similarly for the other $d$, in each case $s(n)$ is even. Since $n$ is doubled at most 3 times, $s(n) \leq \frac{8n}{10}$ .

If $d=4$, then $s(n) = \frac{(10k+4)-4}{10} = k$, i.e. $s(n) < \frac{n}{10}$ .

If $d=5$, then $s(n) = \frac{(10k+5)2}{10} = 2k+1 = \frac{n}{5}$.

If $d=7$, then $s(n) = \frac{(10k+7)2-4}{10} = 2k+1$, i.e. $s(n) < \frac{n}{5}$ .

If $d=9$, we need more steps. Consider $n = 10k+9$ . We double $n$ four times to get a trailing 4, so $s(n) = \frac{(10k+9)16-4}{10} = 16k+14$, i.e. $s(n) < 1.6n$. The last digit of $s(n)$ is $16k+14$ mod $10$, for which we only need to consider $0 \leq k \leq 9$. Trying all $k$ from 0 to 9, we find $s(n)$ ends with $4,0,6,2,8,4,0,6,2,8$ respectively. In particular, $s(n)$ never ends with 9.

Apply the above iteratively. Recall $s(n) < 1.6n$, and $s(n) \mod 10 \in \{0,2,4,6,8\}$ .

If $s(n)$ ends with 0 or 4, then $s(s(n)) \leq \frac{s(n)}{10} < 0.16n$, so $s(s(n)) < n$ .

Otherwise $s(n)$ ends with 2, 6 or 8.
Let $m=s(s(n))$. Then $m$ is even and $m \leq 0.8s(n) < 1.28n$.

If $m$ ends with 0 or 4, then $s(m) \leq \frac{m}{10} < 0.128n$, so $s(m) < n$.
Otherwise $s(m)$ is even and $s(m) \leq 0.8m < 1.024n$, so $s(s(m)) \leq 0.8s(m) < n$.

In every case, there is a sequence of steps taking $n$ to an integer strictly less than $n$, unless we arrive at 4, in which case we've arrived. Call this sequence a jump. Since $n \neq 0$, we never jump to zero.

The jumps reduce $n$ monotonically, so we eventually arrive at a single digit, from which the table above shows that . By inspection of the table above, steps from single digits the sequence always terminates at 4.


  • $\begingroup$ Just as I said to the other answer, I think it should also be proven that "So from then on, the number always decreases". I don't believe this is sufficient proof. Specifically, when $\frac{16n-4}{10}$ ends in an $8$ you are able to reduce the resulting number by doing $\frac{8n-4}{10}$ but that is still larger than the initial $n$ and at this point you haven't proven that this new number doesn't end with a 9. I know it couldn't be but I think you haven't proven that with this alone $\endgroup$
    – Ivo
    Mar 20, 2016 at 14:47
  • $\begingroup$ @IvoBeckers I was assuming the "never ends with 9" comment was understood to automatically invoke the earlier $n \to \frac{8n}{10}$ process. I've now made this explicit. It doesn't matter that the number is higher than the starting $n$, so long as it decreases monotonically at some point. $\endgroup$
    – Lawrence
    Mar 20, 2016 at 14:50
  • 1
    $\begingroup$ What I'm trying to say is. You haven't proven that the not-ending-with-9 process can't produce numbers ending with a 9. For all we know you have a number ending with a 9, gets turned to a number ending in an 8, gets turned to a number with a 9, and so on forever, and this process does not reduce the number $\endgroup$
    – Ivo
    Mar 20, 2016 at 14:55
  • $\begingroup$ @IvoBeckers Ok, let me have a look at this. $\endgroup$
    – Lawrence
    Mar 20, 2016 at 15:00
  • 1
    $\begingroup$ @IvoBeckers The proof has become less elegant, but I think all cases are covered. I wonder whether it would have been simpler to start with 4 and show that we can reach all integers by running the process backwards. $\endgroup$
    – Lawrence
    Mar 20, 2016 at 16:01

Removing rightmost digit is like dividing the number by $10$ for $0$ and more than $10$ for $4$.

If the last digit of the number you have taken is $0$,$4$ you just divide your number by $10$.

If the last digit is $2,7,5$. You multiply your number by $2$ and divide by $10$ that makes dividing by $5$ after $2$ moves.

If the last digit is $1$,$6$ you multiply twice by $2$ then divide by $10$ that makes dividing by $2.5$, still your number gets less.

If the last digit is $3$,$8$ you need to multiply third time by $2$ then divide $10$ that makes dividing by $1.25$, still your number gets lesser and becomes divided by $1.25$ at the end.

The only time your number gets bigger when your last digit is $9$, that makes your number $1.6$ times bigger than before after removing the last digit. But since the next number's last digit (after removing $4$, which becomes last digit after doubling it $4$ times) cannot be $9$ ($x9\times 16$'s second digit cannot be $9$) the mentioned removing number above will make your number to $4$ whatsoever.

So whatever the number you have taken, there will be a solution.

  • $\begingroup$ I came up with the same conclusions as you although I'm still not convinced yet about the 9. It's true that $x9 \cdot 16$ can't have a 9 in the second digit but it could be an 8 and in that case you can only divide by $1.25$ and $1.6 / 1.25$ is still bigger than one and after that you might possibly get another 9 but I'm not sure. I think you still need to expand a bit on the possibilites when the last digit is 9 $\endgroup$
    – Ivo
    Mar 20, 2016 at 14:11
  • $\begingroup$ @IvoBeckers the last digit cannot be 9 again, that's the point. $\endgroup$
    – Oray
    Mar 20, 2016 at 14:19
  • $\begingroup$ No I mean after doing the next step again. Let's say you have a 9 at the end. you multiply by 1.6 and get a 8 at the end. you then can divide by 1.25. At that point your number didn't get smaller and at this point I'm not sure if you could have a 9 again $\endgroup$
    – Ivo
    Mar 20, 2016 at 14:22
  • $\begingroup$ @IvoBeckers it does not matter, i see what u meant but u cannot have 9 as the last digit ever after the last 9 because 94 or 90 as the last 2 digits is impossible after the first removal $\endgroup$
    – Oray
    Mar 20, 2016 at 14:24
  • $\begingroup$ I think you're right but I also think that it should be proven $\endgroup$
    – Ivo
    Mar 20, 2016 at 14:40

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