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For the sake of simplicity, I've been only looking in base $10$ numbers, though I've wondered how this might work in other bases - easier to stick with the familiar, though, since this is purely recreational. Analogous formulations hold in other bases, and are easily constructed (at least for bases $b$ which are integers greater than or equal to $2$).

Insofar as context is concerned, there isn't much other than this was something I was thinking over last night, and I couldn't find any discussions about this online.


Formulation of the Question:

For argument's sake, let us consider our number to be $n$, which can be represented by the following forms:

$$n = ...d_3d_2d_1d_0.d_{-1}d_{-2}d_{-3}... \Leftrightarrow n = \sum_{k \in \mathbb{Z}} d_k 10^k$$

where each $d_k \in \{0, 1, ..., 9 \}$, with its leading and trailing zeroes ignored. (For example, $000012$ is just treated as $12$ and $12.3500000...$ is just treated as $12.35$.) The former consequence means that the first nonzero digit is some number we label $d_f$ with index $f$. So, equivalently, the above forms are given by

$$n = d_f...d_3d_2d_1d_0.d_{-1}d_{-2}d_{-3}... \Leftrightarrow n = \sum_{k = f}^\infty d_k 10^k$$

where each $d_k \in \{0, 1, ..., 9 \}$ and $d_f \neq 0$.

Let us now define a "product of digits," or "digital product," function $dp(n)$ by

$$dp(n) = \prod_{k \leq f} d_k = \text{"the product of n's digits in base 10"}$$

Then the question is as follows:

Is there any "nontrivial" number (at least in base $10$) whose digital product is the number itself, i.e. any natural $n \geq 10$ such that $$dp(n) = n$$

I say trivial because, for single-digit numbers, it is clear that $dp(n) = n$. I'm more interested in "nontrivial results," provided they exist.


Some Thoughts and Investigations:

Ideally, it would be possible to prove $dp(n) \neq n$ for all $n$ of concern, or at least prove there exists some such number(s) that equal their digital product, but I don't think that would be (at the very minimum) an easy task. So I've instead bounced around other ideas and such in my head:

  • There's the one outlined at the end of the previous section, which somewhat motivates me asking this question. That being that, for any single-digit number in any base, $dp(n) = n$.

  • Obviously, too, any number containing any $0$'s satisfies $dp(n) = 0$.

  • Any number between $9^m$ and $10^m$ for a given natural number $m$ will not satisfy $dp(n) = n$, since the maximum digital product for $m$ digits would be $9^m$.

  • Any permutation of the digits of $n$ doesn't change the digital product. Any number with $1$'s in it can also have them safely added or removed without changing the digital product. That is, for example, $2, 21, 121, 121111111$ and any permutations of their digits, all have the same digital product.

  • $n$ must be a whole number. Nuances regarding negatives are not something I want to deal with, so for simplicity this problem is essentially reduced to contending with $n \in \mathbb{N}$. This is because the nonnegative integers are closed under multiplication, and thus $dp(n) \in \mathbb{N} \cup \{0\}$ for all real $n$. (Well, maybe some irrational numbers would have $dp(n) "=" \infty$? That's a question all on its own honestly, but since being irrational would mean having a fractional part, and the fractional part of all digital products is zero, it would still mean $dp(n) \neq n$.)

  • It seems obvious that if $dp(n)$ has less digits than $n$ (or more, but more on that later), then obviously $dp(n) \neq n$. But this begs a certain question: what would be the minimum value for which $n$ and $dp(n)$ would, if not be equal, at least have the same amount of digits? (This question would be motivated by computer searches and could be useful in cutting down computational time further alongside previous observations.)

  • Of course, this question could also be framed in another way, though my mathematical background is less certain where this comes from. Namely, consider repeated application of the digital product function, i.e. $dp(n), dp(dp(n)), dp(dp(dp(n))),...$. The alternative framing of this question in that light would be that, is there any number with two or more digits such that under iteration the function eventually cycles (getting stuck in a loop) without reaching single digit numbers?

  • On research into a related topic, sum-product numbers, it was discussed on MSE that $dp(n) \leq n$. (Our investigation is concerned with where they're equal.) So in that sense, on repeat iteration, the digital product is decreases monotonically (albeit not strictly monotonically). That is,

$$ \text{one of 0-9} ... \leq dp^{i+1}(n) \leq dp^i(n) \leq dp^{i-1}(n) \leq ... dp^2(n) \leq dp(n) \leq n$$


A Crude Brute Force Search:

I have made a relatively crude MATLAB script that (as posted) checks numbers from $10$ to $10^{10}$ for $dp(n) = n$. (Crude in that it doesn't really incorporate any of the observations above, just checks all of them. I'm really, really terrible with programming.)

Currently, it's checked all those up to $5 \times 10^{9}$ and none (that have more than one digit) have met that condition. This suggests that, if such an $n$ does exist, it will be relatively large, though given how slowly $dp(n)$ obviously grows over time it probably doesn't exist (though proving that eludes me - this is a bit above my head).

I doubt I can check many more numbers, however. I'm running this script on my laptop which is quite out of date and is just a really poor machine overall. The script itself could probably be refined as well.


So My Questions To You:

  • Either by brute force or some other means, can you find a (nontrivial) number such that $dp(n) = n$?

  • I'm currently at the point of conjecturing no such $n$ exists, but I have no clue how one might prove that. (Like, legitimately no clue. I have no idea where to begin for something like this.)

  • I noticed that the number of sum-product numbers is finite. Based on their definition, perhaps it might suggest likewise that the number of numbers such that $dp(n) = n$ is also finite?

  • Perhaps just any noteworthy thoughts/findings on the matter, or research into this that I overlooked?

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    $\begingroup$ For $n = 10a + b$, $a\cdot b < n$ because given $0<=a,b<=9$, Let $a$ be any number. For $n$, it is multiplied by 10. $b$ will never be greater than $9$ so $10a > b\cdot a$, therefore $n > a\cdot b$. This may be applied to other bases and for higher place values, but this is just a bit of thinking to chew on. Good luck! $\endgroup$ – Christopher Marley Dec 27 '18 at 6:21
  • $\begingroup$ Why did you bother with negative powers of $10$? The product of integers is an integer so no non-integer need be considered. $\endgroup$ – fleablood Dec 31 '18 at 2:27
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Your conjecture is true. If it isn't, let us suppose that there is a number $m>9$ of $n$ digits such that $dp(m)=m$. As you have checked and also proved by @ChristopherMarley in the comments, $n>2$.

Let's chop off the last digit of $m$ to get $m'$ of $n-1$ digits. Now, $m' \leq m/10$ (since we're taking the floor of $m/10$) and $dp(m') \geq m/9$(since the last digit is $9$ at most) $> m/10$.

Hence, $m' \leq dp(m')$. Continuing this chopping off of last digit, we'll get a number $m_0$ of $2$ digits such that $m_0 \leq dp(m_0)$ but we know that's not possible.

So, such an $m$ doesn't exist i.e. your conjecture is true that single digit numbers are the only numbers with $dp(m)=m$.

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    $\begingroup$ Wow, that was a wonderfully simple and elegant proof. I figured it'd be much more convoluted. I love it. ^_^b $\endgroup$ – Eevee Trainer Dec 27 '18 at 7:23
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    $\begingroup$ @EeveeTrainer I thought so too. I was trying to improve on your Matlab script when it struck me. Haha. $\endgroup$ – SinTan1729 Dec 27 '18 at 7:24
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As you've observed, it's not hard to see that there is no such number with more than one digit. Considering three-digit numbers, just for example, we see that for a number of the form abc to have the desired property, we would need $a\cdot b\cdot c = 100a+10b+c$. The left side is at most $81a$ and the right side is at least $100a$, so they can't be equal. A similar argument works for any numerals of more than a single digit.

But this argument depends on the fact that no digit exceeds 9. If we are willing to relax the rules about base-10 numerals a bit, and allow digits larger than 9, we can find solutions. So for example when we count “seventy-seven, seventy-eight, seventy-nine” instead of continuing with eighty we are allowed to continue “seventy-ten, seventy-eleven, seventy-twelve…”, which we will write 7 7, 7 8, 7 9, 7 10, 7 11, 7 12

What does a number like “seventy-twelve” mean, anyway? Simple: Just as “seventy-nine” means seven tens and nine units, or $7·10+9 = 79$, “seventy-twelve” means seven tens and twelve units, or $7·10+12 = 82$. 7 12 is just another way to write 8 2. They represent the same number, similar to how $\frac12$ and $\frac48$ are also two different ways to write the same number. In the conventional decimal system there is only one way to write each number. But in this relaxed system, there may be several ways. For example, the number $121$ can be written as 1 2 1 or as 12 1 or as 11 11 or even as 9 31. (That last is nine tens and 31 units, or $9\cdot 10 + 31$.)

Now there are some numbers that are equal to their digital product. For example, 6 12 (“sixty-twelve”) is another way to write $72$, and indeed $6\cdot 12=72$!

Simple algebra ($10a+b = ab$, so we want $a = \frac b{b-10}$) finds the two-digit examples:

\begin{array}{rlr} \text{Relaxed numeral} & \text{Silly name} & \text{Conventional numeral} \\ 11\ 11 & \text{(eleventy-eleven)} & 121\\ 6\ 12 & \text{(sixty-twelve)} & 72 \\ 3\ 15 & \text{(thirty-fifteen)} & 45 \\ 2\ 20 & \text{(twenty-twenty)} & 40 \end{array}

And of course there are solutions with more digits, such as 6 16 8 (six hundred sixteenty-eight) $= 768$.

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  • $\begingroup$ I'd consider this more an "exploit" because the digits in a number are not to exceed the base. $\endgroup$ – iBug Dec 27 '18 at 14:21
  • $\begingroup$ True, it's a bit of an exploit, but it's still pretty neat to think about how things would be if we loosened conventions, you know? $\endgroup$ – Eevee Trainer Dec 28 '18 at 0:05
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Note: $dp(N)$ is an integer. So if $dp(N) = N$ then $N$ is an integer.

If $N = \sum\limits_{k=0}^n d_k10^k$ has more than one digit, i.e. $n \ge 1$ and $N$ is such a number then

$d_0d_1.....d_{n-1}d_n =\sum\limits_{k=0}^n d_k10^k > d_n 10^n$

$d_0d_1.....d_{n-1} > 10^n$

$9^n \ge d_0d_1.....d_{n-1} > 10^n$

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