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Call the first digit of a number digit $0$. The digit after that digit $1$, and so on and so forth.

In base $10$, the number $6210001000$ describes itself, because digit $0$ is $6$ and it has six $0$s. Digit $1$ is $2$ and it has $2$ $1s$. Digit $2$ is $1$ and there is only one $2$. There are none of the other numerals.

So it occurs to me that in duodecimal, you can do $821000001000$, which is $6073061476032$ in decimal. In base $40,$ you could do $Z21000...$ something like that, you get the idea. You could keep going, the only problem being not having enough symbols.

According to a graphic I saw on Facebook, $6210001000$ is the only such number in base $10$. I'm wondering: might there be some base in which a number describes itself like this but its first digit isn't 4 less than the base, followed by $21$ and a bunch of 0s with a 1 close to the end?

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For bases sufficiently large (as @Goos noted, there are examples for low bases), it is impossible. I will use the facts that there must be as many numbers as the base, and that the digits must sum to the base (since they count the digits in the number).

Take the base $=n$, and consider the first digit $k$. $k$ cannot be zero, because that would say there are zero zero's, which there wouldn't be! So there are $k$ zero's, and at least one $k$ (for now), meaning we have $$n-k-1$$ digits left to fill. These digits must add to the remaining $n-k-k\cdot 0=n-k$, and the only $n-k-1$ digits which sum to $n-k$ are $$1,1,\ldots,1,2$$ (we have $n-k-2$ one's). I will eliminate $k=1,2$ later. So we have $2$ of something, not $k$'s and not zero's. We must have $2$ one's. And then we have one $k$ and one $2$, which works. The condition that the numbers sum to the base gives $$k+1+1+2+k\cdot 0=k+4=n \implies k=n-4$$ so the only possible self-descriptive number is $$(n-4)(2)(1)0\ldots(1)0\ldots0.$$


If $k=1$, then there will have to be two one's, and one two, and one zero. This gives the only possibility $$(1210)_4$$ in base $4$.

If $k=2$, then there must be two two's. We have two zero's, so there are one or no one's. The two possibilities here are $$(21200)_5,\quad (2020)_4$$ So we have enumerated all self-descriptive numbers.

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Yes, there are, but it's not known if there are a lot of them.

Let me rephrase "the pattern 6210001000" algebraically like so: $(b - 4)b^{b - 1} + 2b^{b - 2} + b^{b - 3} + b^3$. So the smallest base in which this results in a digit pattern similar to 6210001000 is base 7. But since $6^5 = 7776$, maybe we can do an exhaustive search in binary, ternary, quartal, quintal and heximal.

This turns up 1210 in quartal, which is 100 in decimal, 2020 in quartal, which is 136 in decimal, and 21200 which is 1425 in quintal. None of these are 292 nor 901, so they are not of the described pattern. There are apparently none in base 6.

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    $\begingroup$ Very good. Only thing is I'm wondering if you didn't mean $6^6 - 1$ rather than $6^5$. Actually, there is one other detail, but I think it makes more sense for me to fix it for you... $\endgroup$ – Robert Soupe Jun 28 '14 at 3:30
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Not a complete answer to your question, but I believe that the problem you're describing can be reduced to the following:

Given a natural number $B>5$, prove that there is only a single multi-set $M$ of natural numbers, such that $\sum{M}=\prod{M}$ and $|M|=B-4$.

For example:

  • $B=10$
  • $M=\{6,2,1,1,1,1\}$
  • $\sum{M}=\prod{M}$ and $|M|=B-4$

If there is more than a single such multi-set, then there are numbers that describe themselves in some base but not according to the illustrated pattern.

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  • $\begingroup$ Could you elaborate on how you reduced the problem to this? $\endgroup$ – punctured dusk Aug 7 '14 at 9:49
  • $\begingroup$ @barto: See the additional note at the bottom of the answer. $\endgroup$ – barak manos Aug 7 '14 at 9:56
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From what I understand, the number 10,213,223 qualifies. It has one zero, two ones, three twos, and two threes. That's what led me to Google this idea of numbers that describe themselves. I'm trying to see what other numbers qualify.

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  • $\begingroup$ That technically counts because I was not specific enough. But if I build it from your description, I get 1233000000, which is clearly not the same as 10213223. $\endgroup$ – user156970 Nov 19 '14 at 22:05

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