Ponder This Challenge - January 2026 - Number splitting

The decimal notation string of the number $123$ can be split in various ways: To $12$ and $3$ , to $1$ and $23$, to $1$, $2$ and $3$, and to $123$ . For each way to split the number, we add up the obtained parts. For $12$ and $3$ , we obtain the number $15$; for $1$ and $23$, we obtain the number $24$; for $1$, $2$, $3$ we obtain $6$; and for $123$ , we obtain $123$ itself. We denote this by:

For a general natural number $n$ , we define $A_n$ similarly as the set of all natural numbers that can be obtained by splitting and adding the decimal notation of $n$ . For example:

Given a set $A$ of integers and an integer $n$ we use the standard notation $n\cdot A\triangleq\{n\cdot x\ |\ x\in A\}$. (where · is the usual multiplication of numbers).

Your goal: Find the sum of all the natural numbers $x$ such that there exists $1\le n\le 10^{6}$ for which $x\in n\cdot A_n$ and $n\in A_x$.

A bonus "*" will be given for finding the sum of all the natural numbers $x$ such that there exists $1\le n\le 10^{7}$ for which $x\in n\cdot A_n$ and $n\in A_x$.

Solution

• Click here to view the solution

  The numerical solutions are

  160808197419276
  26190672886645170
  

  The solutions can be found by brute-force over all values of $n$ in the range, but there are nice optimizations to be found. Here is a simple an effective one suggestd by Todd Will:

  First, note that since $10\equiv 1(\text{mod}\ 9)$, the sum of digits (and groups of consecutive digits) of $x$ is equivalent modulo 10 to $x$ itself. Hence, if $n\in A_x$ we have $n\equiv x(\text{mod}\ 9)$.

  Similarily, $x\in n\cdot A_n$ implies $x\equiv n\cdot n(\text{mod}\ 9)$, so we obtain $n\equiv n^2(\text{mod}\ 9)$ meaning $n=9k$ or $n=9k+1$, cutting down the search space.

Solvers

• *Alper Halbutogullari (30/12/2025 10:54 AM IDT)
• *Nadir S. (30/12/2025 11:20 AM IDT)
• *Lazar Ilic (30/12/2025 11:33 AM IDT)
• Alex Fleischer (30/12/2025 12:57 PM IDT)
• *Bertram Felgenhauer (30/12/2025 1:05 PM IDT)
• *Ahmet Yüksel (30/12/2025 3:11 PM IDT)
• *Ankit Aggarwal (30/12/2025 3:23 PM IDT)
• Abraham AJ Arshad (30/12/2025 3:58 PM IDT)
• *Paul Lupascu (30/12/2025 6:02 PM IDT)
• *Dan Dima (30/12/2025 6:38 PM IDT)
• *Prashant Wankhede (30/12/2025 9:08 PM IDT)
• *George Jiri Spitalsky (30/12/2025 10:14 PM IDT)
• *Reiner Martin (31/12/2025 12:19 AM IDT)
• *Jean-François Hermant (31/12/2025 1:56 AM IDT)
• Loukas Sidiropoulos (31/12/2025 2:14 AM IDT)
• *Sergey Batalov (31/12/2025 2:58 AM IDT)
• *King Pig (31/12/2025 3:37 AM IDT)
• *Guangxi Liu (31/12/2025 10:29 AM IDT)
• *Bert Dobbelaere (31/12/2025 12:42 PM IDT)
• Michael Branicky (31/12/2025 11:17 PM IDT)
• Gary M. Gerken (31/12/2025 11:57 PM IDT)
• *Asaf Zimmerman (1/1/2026 7:04 PM IDT)
• *Daniel Bitin (1/1/2026 7:05 PM IDT)
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• David Keyes (4/1/2026 1:03 AM IDT)
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• *Teerachot Saenthong (5/1/2026 9:20 AM IDT)
• *Stéphane Higueret (5/1/2026 12:08 PM IDT)
• *Vladimir Volevich (5/1/2026 3:13 PM IDT)
• *Kandala Savitha Viswanadh (5/1/2026 6:22 PM IDT)
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• *Peter Moser (5/1/2026 9:41 PM IDT)
• *Florian Sikora (6/1/2026 12:08 AM IDT)
• *Daniel Chong Jyh Tar (7/1/2026 5:50 PM IDT)
• *Ankit Chakraborty (7/1/2026 11:00 PM IDT)
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• *Michael Liepelt (8/1/2026 9:27 PM IDT)
• *Justin Mazenauer (9/1/2026 4:08 PM IDT)
• *Todd Will (10/1/2026 8:50 PM IDT)
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• *Hansraj Nahata & Govind Jujare (19/1/2026 7:42 PM IDT)
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• *Isabelle Mirouze (21/1/2026 11:58 AM IDT)
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