Recall from ŧhis question that we call a positive integer slimdownable or slim for short if it is part of a sequence of integers where each is followed by itself divided by its length, i.e. its number of digits. In particular, each must be divisible by its length and the sequence will be falling until it hits a single-digit number.


108: slim because $108\overset{/3}{\rightarrow}36\overset{/2}{\rightarrow}18\overset{/2}{\rightarrow}9\overset{/1}{\circlearrowleft}$

78: not slim because $78\overset{/2}{\rightarrow}39\overset{/2}{\rightarrow}\Vert$

Prove or disprove that for any positive integer $n$ there exists a slim number with $n$ digits.

Note: you may use a computer either to produce a counter example or, for example, to complement an asymptotic result, or whatever else you see fit. If you choose to do so, in order to validate your code please answer the following test questions:

Are there any solutions with $11111$ digits? If yes: How many? What are the first 10 digits of their median?
For each $n$ between $1$ and $11111$ calculate the number od solutions with $n$ digits. What is the largest count?

Here is a test case which you can use as a quick sanity check for your code:

At 3590 digits there are four solutions. They all slim down to $6$. First ten digits of these numbers are '3159252337...', '3735860235...', '4606981484...', '6706597705...'.

  • $\begingroup$ Shouldn't this be on maths.se? $\endgroup$
    – Caius Jard
    Commented Sep 10, 2020 at 16:39
  • 4
    $\begingroup$ It's not serious math, it's recreational math. There are lots of this kind of question here. $\endgroup$ Commented Sep 10, 2020 at 16:42
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    $\begingroup$ Yeah, this fits right in. $\endgroup$
    – Bass
    Commented Sep 10, 2020 at 16:55
  • $\begingroup$ Didn't you just disprove it with your example of 78 above? $\endgroup$
    – APrough
    Commented Sep 10, 2020 at 16:55
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    $\begingroup$ @APrough there are at least twelve other 2-digit numbers though. $\endgroup$
    – Bass
    Commented Sep 10, 2020 at 16:57

1 Answer 1



For any $n$-digit number, there cannot exist a slim number for all positive integer $n$.


For all $n$, there must be a multiple of $n$ that is $n$-digits long.

Let's call our $n$-digit number $Z$.
Take into account that in $10^{ceil(log(n))}$ - let's call this $a$, there must be $a/n$ multiples of $n$. This provides our way down the slimming process.
Let's look back at $Z$. By the preceding 2 steps, provided we chose the right $Z$, there will be at least one number $y$ of length $(n-a)$ or $(n-a+1)$ where $y = Z/n$ and can be expressed with a new $Z$ to repeat the process.

I am now the proud owner of an Excel spreadsheet that generated new values starting from 5-9 (since 1-4 cannot generate a 2-digit number) and ending where Excel hits a #Value error.

The image below is in descending order (9 to 5)


What's eating away at me is large $n$.

For $n$ = 1 billion, I can't guarantee that there will be a slim number for $n$ between 1 billion - 1 and 1 billion - 9, especially since I only have 5 starting paths.
And since the gap would only get wider and wider as $n\rightarrow\infty$, there can't be a slim number for all positive integers $n$.

e.g. For $n$ = $1E1,000,000$, my division would cause the first number to drop by 6 places, and I might be able to populate the gap with only 5 deterministic paths, but eventually, it wouldn't be filled.

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    $\begingroup$ You are thinking along the right lines, but there is at least one inaccuracy: To go upward you have to multiply with t he length of the larger number, not the smaller. For example, starting from 28: downwards no problem: /2 -> 14 ; /2 ->7. But upwards first step no problem: 28; *2 -> 56 but now you have to do *3 not *2 ->168 etc. $\endgroup$ Commented Sep 11, 2020 at 2:39
  • $\begingroup$ @PaulPanzer Right, thanks for pointing that out. I ran across that when trying to check, but forget when I typed it in. The method that I typed up does it correctly, though the logic might not quite be there. $\endgroup$ Commented Sep 11, 2020 at 3:54

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