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This question asks for nonzero numbers whose values match the average of their own digits. Such numbers are not immediatelly evident by intuition, but it is not difficult to show that there are in fact infinitely many solutions.

In one answer a solution is constructed by the following method: Start with a finite representation lying between a "low digit" and a "high digit", then append either the low digit or the high digit to drive the average towards the evolving representation. Because of the exponentially decreasing place values of the successively appended digits, the representation is stable enough to assure convergence with the truncation error tending to zero. Because any finite digital represesentation within range can be used as the initial value, a countable infinity of solutions is identified by this method.

In base three or higher, we can modify this procedure by allowing a choice of two or more values for the high digit, the low digit or both. Since high and low digits must persistently alternate to reach the required equality, this modification multiplies the cardinality of solutions by $2^\alpha$ where $\alpha=\aleph_0$, so the countable infinity is promoted to an uncountable one.

But this promotion does not work in base two, where perforce the low digit must be 0 and tge high digit must be 1. So the method described here renders only the countable infinity.

Question: Is there a construction method that gives an uncountable set of solutions in base two? Or, is the solution set in that base inherently only countably infinite?

Oscar Lanzi
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  • Are these integers? There are only countably many integers, no matter what the base for the representation is. Am I misunderstanding? You're looking for a collection of integers with some property? – MPW Jun 05 '25 at 18:07
  • Are you asking if for every number $x \in [0, 1]$ there is continuum solutions of $\theta_2(y) = x$? – mihaild Jun 05 '25 at 18:24
  • @mihaild The question is about the cardinality of ${x \in \mathbb{R} : \theta(x) = x}$. (Of course, such an $x$ lies in $[0,1]$.) – Dermot Craddock Jun 05 '25 at 18:29
  • I said uncountable infinity. The difference is seen by considering the Cantor set. Uncountable solutions in higher bases would look more like the Cantor set than a continuum. – Oscar Lanzi Jun 05 '25 at 18:31
  • @OscarLanzi sorry, I don't understand - Cantor set has cardinality continuum. Anyway, can't you for numbers in $(1/3, 2/3)$ use the same method, but add digits 3-by-3 instead of 1-by-1 - $001$ or $010$ for low "digit", $110$ or $101$ for high? – mihaild Jun 05 '25 at 18:37
  • @mihalid Possibly. Maybe this is an answer? – Oscar Lanzi Jun 05 '25 at 18:39
  • I am thinking "continuum" means you fill 8n an interval, which the Cantor set does not. Am I not using the righr definitiin in these comments? – Oscar Lanzi Jun 05 '25 at 18:41
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    Ah, sorry, my favorite homonym. There is continuum in topology (compact connected metric space) and in set theory (cardinality of real numbers). It seems you thought about the first one and I saw cardinality and thought abouth the second. – mihaild Jun 05 '25 at 18:42

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We can use the same method, but with triplets of digits instead of single digits (triplets instead of pairs so we can avoid dyadic rationals with non-unique representation).

Start with $x_0 = 0.100_2$. We will built a unique solution for each $S \subseteq \mathbb N$, building a monotonically increasing sequence $x^S_n$ s.t. $x^S_n \in (1/3, 2/3)$, $\theta(x^S_n) - x^S_n \to 0$ (so $x^S_n \to x^S$ which is solution), and $S$ and $S'$ first time differ in inclusion of number $i$ (one of them has it and other doesn't), then $x^S$ and $x^{S'}$ differ in $3i+3$-rd digit.

On $n$-th step:

  • If $\theta(x_n) < x_n$ and $n \in S$, append $110$
  • If $\theta(x_n) < x_n$ and $n \not\in S$, append $101$
  • If $\theta(x_n) \geq x_n$ and $n \in S$, append $001$
  • If $\theta(x_n) \geq x_n$ and $n \not\in S$, append $010$
mihaild
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