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How to Determine Numbers That Preserve Digit Composition When Multiplied

Mathother
Explained on January 11, 2026
📚 Grade graduate🟡 Medium⏱️ 10 min

Problem

Is there a non-negative integer with this property when multiplying? The property in question is that when multiplied with a number containing only $0$s and $1$s (in base ten), the result also only contains $0$s and $1$s, but that is not true when multiplied by the number's palindrome. (Eg. the number $X \cdot 1011$ produces only $0$s and $1$s, but $X \cdot 1101$ does not). A clue I've been given is that no numbers with less than ten digits have this property, so I think it has to do with carry-over. If no such numbers exist, a proof of why would be very helpful. To be clear, the other input number must always be purely made of $0$s and $1$s, and the output number must also be for the base number, but must not be for the palindrome of the other input number.

Step-by-Step Explanation

Hello! This is a fascinating problem that combines number theory with pattern recognition. Let me help you work through this step by step.

What We're Solving:

We need to find if there exists a non-negative integer X such that:
  • X × (some binary-looking number) = result with only 0s and 1s
  • X × (palindrome of that binary-looking number) ≠ result with only 0s and 1s

The Approach:

This problem requires us to think about what makes multiplication produce only 0s and 1s as digits. The key insight is understanding when carry operations occur in multiplication and how palindromes might disrupt favorable patterns.

Step-by-Step Solution:

Step 1: Understanding the constraint First, let's think about what it means for a product to contain only 0s and 1s. This is quite restrictive! When we multiply X by a number made of 0s and 1s, we're essentially adding shifted copies of X (where we encounter 1s) and adding nothing (where we encounter 0s).

Step 2: Analyzing the carry problem The clue mentions numbers need at least 10 digits, suggesting carries are crucial. When we multiply, carries occur when intermediate sums exceed 9. For the result to have only 0s and 1s, we need very careful control over these carries.

Step 3: Consider what palindromes change When we flip a binary-looking number to its palindrome, we change the positions where 1s appear. This means we're adding shifted copies of X in different positions. If X was carefully chosen to work with one pattern, the palindrome might cause problematic carries.

Step 4: Exploring small cases Let's think about why smaller numbers don't work. With fewer digits, there are fewer opportunities to create the delicate balance needed between the original number and its palindrome having different carry behaviors.

Step 5: The theoretical approach For such an X to exist, we need:

  • X has a special structure that, when multiplied by certain patterns of 0s and 1s, produces controlled carries
  • The palindrome version disrupts this careful balance
Step 6: Research direction This appears to be related to the study of "digital restrictions" in number theory. The existence of such numbers is still an open question in mathematics!

The Answer:

This is actually an unsolved problem in number theory! No such number has been found yet, but it also hasn't been proven that none exists. The problem touches on deep questions about digit patterns in multiplication and the behavior of carries in base-10 arithmetic.

The search continues for such numbers, with computer searches having checked very large ranges without success, but a complete proof of non-existence remains elusive.

Memory Tip:

Think of this as the "palindrome disruption problem" - we're looking for a number that plays nicely with one pattern but gets "disrupted" when that pattern is flipped like a mirror!

This is a beautiful example of how simple-sounding questions can lead to deep mathematical mysteries. Keep exploring - you're thinking about cutting-edge number theory!

This explanation was generated by AI. While we work hard to be accurate, mistakes can happen! Always double-check important answers with your teacher or textbook.

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📷 Problem detected:

Solve: 2x + 5 = 13

Step 1:

Subtract 5 from both sides...

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