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I've been playing around with an idea about composite numbers and the digits of their factors. I've noticed a certain pattern, and for lack of a better term, I've started calling numbers that exhibit this pattern digit-disjoint." (I haven't found this term used elsewhere, so I think I might be coining it). A number is digit-disjoint if:

  • $n$ is odd composite number which is greater than $3$.
  • $n$ can represented in at least two $x \times y$ forms where none of $x$ or $y$ is 1
  • None of their $x \times y$ representations use any digit from $n$.

For example, the number 81 is digit-disjoint. Its digits are $8$ and $1$. It can be factored as $9 \times 9$ and $3 \times 27$, which none of these use $8$ or $1$.

I stumbled upon this idea while using a small tool I wrote to find factorizations of numbers. I've started exploring it and have found a few examples.

  • I wrote a program and searched all numbers from $1$ to $50,000,000$ and found some these numbers. I also edited the program to only search numbers only using $4$ and $7$ or $8$ and $1$ digits, allowing to search much bigger numbers. Biggest number found so far was $477,474,774,477$, can represented in $159 \times 3002986003$, $53 \times 9008958009$ and $3 \times 159158258159$.

  • It seems like these numbers get rarer as they get bigger, but they don't seem to stop.

Every number could represented in $1 \times n$, but the $1$ factor here is ignored.

So, my main question is: are there infinitely many of these numbers, or do they eventually stop?

I'm also curious about their density. Does the proportion of digit-disjoint numbers among all odd composites get closer and closer to zero as the numbers grow? Any thoughts on how to even approach constructing more of them would be amazing.

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    $\begingroup$ Probably not much studied about this, because mathematicians generally don't care much about digital representations. Certainly more of a recreational mathematics problem. $\endgroup$ Commented 5 hours ago
  • $\begingroup$ This is interesting. The only result that comes to mind here is from Maynard, 2019, which, among other things, shows that there are infinitely many primes without the digit $7$. Can that be used here? If you loosen the constraint from two $x\cdot y$ forms to $1$, maybe. $\endgroup$ Commented 3 hours ago

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Sadly, there's a "dumb" answer. Let $n_k$ be the number that is $k$ copies of the digit $n$; e.g., $3_4 = 3333$. To resolve any ambiguity, let $n_0$ be an empty list of digits. Let $a_b c_d$ be notation for $b$ copies of the digit $a$ followed by $d$ copies of the digit $c$; e.g., $1_2 3_4 = 113333$.

Then $$ 3_k \cdot 3_k = 1_{k-1}0_18_{k-1}9_1 \text{.} $$ Examples: \begin{align*} 3 \cdot 3 &= 1_0 0_1 8_0 9_1 = 9 \\ 33 \cdot 33 &= 1_1 0_1 8_1 9_1 = 1089 \\ 333 \cdot 333 &= 1_2 0_1 8_2 9_1 = 110889 \\ &\vdots \end{align*}

It's a fairly straightforward induction to show that this continues forever.

There's a similar pattern for $6_k \cdot 7_k$, for $k \not\in \{2, 5, 8, 11, \dots\}$, the integers congruent to $2 \pmod{3}$. And all of $6_k \cdot 3_k$. Each of these is exponentially growing, so the density of examples of this type goes to zero, but as you show there are examples not of this type, so the density might not asymptotically vanish.

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