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That's surprising. Any informal thoughts why would even a single 4-digit constant exist with this property? The intuition would be there are multiple cycles in

That's surprising. Any informal thoughts why would even a single 4-digit constant exist with this property? The intuition would be there are multiple cycles in this graph.

Arnavion 3y ago
The first reference in the article has the working out. https://plus.maths.org/content/mysterious-number-6174 https://plus.maths.org/content/mysterious-number-6174
- dcow 3y ago
  The conclusion is that we don’t know and therefore it might be coincidental. Hardly satisfying. It does appear there are cycles for other lengths.
  - layer8 3y ago
    When does a mathematical fact become a coincidence?
    - v64 3y ago
      Relevant reading: What Are Mathematical Coincidences (and Why Does It Matter)? by Marc Lange [1] [1] https://philosophy.unc.edu/wp-content/uploads/sites/122/2013/10/mathcoincMINDpublished.pdf https://philosophy.unc.edu/wp-content/uploads/sites/122/2013...
      - cubefox 3y ago
        The term "relevant" might be an understatement here.
    - noah_buddy 3y ago
      Layman’s guess: if there’s no generalizable proof describing why this works and the only method of proof is enunerating every case. That said, I kinda doubt that that is the case here, we probably just need a craft way of describing this so that we may come up with a crafty way of proving it :)
      - SkyBelow 3y ago
        If there is a proof of no generalizable proof, that is interesting in and of itself. If there is not, then doesn't that mean we haven't found it yet (or proven it impossible, or proven it impossible to prove if impossible...). If we are still at the point of enumerating each case, that means our knowledge of the subject matter is lacking the deeper understanding to solve it in a nicer fashion.
  - AnotherGoodName 3y ago
    Not an answer but a very good clue I can see is that the sum of the digits on each side of the equation consistently gets closer and closer. 9541 – 1459 = 8082 Left hand digit sum = 19. Right = 18. They are 1 apart. 8820 – 0288 = 8532 Both sides now = 18. Now 0 apart and they'll stay there. They are only 0 apart when at 18. 8532 – 2358 = 6174 Both sides = 18 7641 – 1467 = 6174 Both sides = 18 You can play with this a bit and it's consistent. The sum of digits of the left and right hand side consistency get closer to each other iteratively (but not necessarily closer to 18). Eventually they lock in at being equal to each other when their digits sum to 18. This seems to be one property to look at. I think there's then a second thing happening. Once the values on both sides have digits that sum to 18 the process from there converges on to 6174. So first the digits of the two sides to the equation converge to equal the same which always only occurs when the digit sum is 18. The digit sum locks into being at 18 at that point. And then subsequently once the digits are 18 they converge on to 6174. I would start by working out why digits on each side of the equation converge to summing to 18 on both sides of the equation and never being equal at any other value in this process. It reminds me of https://math.stackexchange.com/questions/99725/every-integer-is-congruent-to-the-sum-of-its-digits-mod-9 https://math.stackexchange.com/questions/99725/every-integer... Now the next thing I would do is ask why does every number with digits that sum to 18 eventually end up at 6174. 4 digit numbers with digits that sum to 18 is a very limited set so it should be easy to figure out the combinations and why they all reach 6174. Put those two together and you'd have an answer. (I'm thinking about it now but it really doesn't seem too hard).
  - vanderZwan 3y ago
    The follow-up article appears to give some hints though: > A number of readers emailed to say they had discovered that repeatedly adding up the digits of any of the kernels of Kaprekar's operation always equalled 9 (...) Professor Nishiyama has provided an explanation why this happens: it is because the result of performing Kaprekar's operation on any number is a multiple of 9. https://plus.maths.org/content/pluschat-15 https://plus.maths.org/content/pluschat-15 This is starting to look very similar to "if you repeatedly add all the digits of an integer represented in base 10 representation until you have a single digit, and the result is 3, 6 or 9, then it is divisible by 3". I forgot the exact explanation for that one, but IIRC has to do with the implicit calculation that is embedded in base 10 positional notation, other bases have a different number you can quickly verify the divisibility of this way. So maybe that (the "implicit calculation in base 10 representation" thing) is one part of the explanation. I mean whatever it is, it feels like a mix of all these operations imposing constraints upon each other and interacting with the recursive feedback loop to result in the convergence as an emergent property.
    - Arnavion 3y ago
      >This is starting to look very similar to "if you repeatedly add all the digits of an integer represented in base 10 representation until you have a single digit, and the result is 3, 6 or 9, then it is divisible by 3". I forgot the exact explanation for that one, but IIRC has to do with the implicit calculation that is embedded in base 10 positional notation Yes. The number "xyz" is 100x + 10y + z. Each power of 10 can be split into 1 plus a multiple of 9, ie (x + y + z) + (99x + 9y). The second group where all the components are a multiple of 9 is of course divisible by 9 and by 3. The first group is the sum of the digits. So if the sum of the digits is divisible by 3, then the original number was divisible by 3. If the sum of the digits is also divisible by 9, then the original number was also divisible by 9. If the sum of the digits is not divisible by 3, the original number was not divisible by 3 either. This generalizes to any number of digits, and to any base N for testing divisibility by N-1 or factors of N-1.
t-3 3y ago
Check out https://en.wikipedia.org/wiki/Kaprekar%27s_routine https://en.wikipedia.org/wiki/Kaprekar%27s_routine At a glance, there seem to be some patterns, like how for those bases with a 2-digit Kaprekar number the sum of the digits is base-1. There must be some number theory explanation for it.
bmacho 3y ago
The graph is small. Similar graphs (e.g. this algorithm for 2 or 5 digit, or 3-4 digit numbers for base 9 or 11) have multiple cycles: https://en.wikipedia.org/wiki/Kaprekar's_routine#Kaprekar's_constants_and_cycles_of_the_Kaprekar_mapping_for_specific_base_b https://en.wikipedia.org/wiki/Kaprekar's_routine#Kaprekar's_...
penteract 3y ago
One thing that makes it less surprising is that there are lots of numbers which map to the same result - for example all permutation of a bag of digits. I checked, and there are only 55 distinct results (54 excluding 0000) from applying the process to all 4-digit numbers, which leaves less space for lots of cycles.
underlipton 3y ago
Does it work for different base systems? As someone who failed high school calculus (i.e., a moron), I would imagine that it's an artifact of base-10.

That's surprising. Any informal thoughts why would even a single 4-digit constant exist with this property? The intuition would be there are multiple cycles in this graph.

Arnavion 3y ago

The first reference in the article has the working out. https://plus.maths.org/content/mysterious-number-6174 https://plus.maths.org/content/mysterious-number-6174

dcow 3y ago
The conclusion is that we don’t know and therefore it might be coincidental. Hardly satisfying. It does appear there are cycles for other lengths.
- layer8 3y ago
  When does a mathematical fact become a coincidence?
  - v64 3y ago
    Relevant reading: What Are Mathematical Coincidences (and Why Does It Matter)? by Marc Lange [1] [1] https://philosophy.unc.edu/wp-content/uploads/sites/122/2013/10/mathcoincMINDpublished.pdf https://philosophy.unc.edu/wp-content/uploads/sites/122/2013...
    - cubefox 3y ago
      The term "relevant" might be an understatement here.
  - noah_buddy 3y ago
    Layman’s guess: if there’s no generalizable proof describing why this works and the only method of proof is enunerating every case. That said, I kinda doubt that that is the case here, we probably just need a craft way of describing this so that we may come up with a crafty way of proving it :)
    - SkyBelow 3y ago
      If there is a proof of no generalizable proof, that is interesting in and of itself. If there is not, then doesn't that mean we haven't found it yet (or proven it impossible, or proven it impossible to prove if impossible...). If we are still at the point of enumerating each case, that means our knowledge of the subject matter is lacking the deeper understanding to solve it in a nicer fashion.
- AnotherGoodName 3y ago
  Not an answer but a very good clue I can see is that the sum of the digits on each side of the equation consistently gets closer and closer. 9541 – 1459 = 8082 Left hand digit sum = 19. Right = 18. They are 1 apart. 8820 – 0288 = 8532 Both sides now = 18. Now 0 apart and they'll stay there. They are only 0 apart when at 18. 8532 – 2358 = 6174 Both sides = 18 7641 – 1467 = 6174 Both sides = 18 You can play with this a bit and it's consistent. The sum of digits of the left and right hand side consistency get closer to each other iteratively (but not necessarily closer to 18). Eventually they lock in at being equal to each other when their digits sum to 18. This seems to be one property to look at. I think there's then a second thing happening. Once the values on both sides have digits that sum to 18 the process from there converges on to 6174. So first the digits of the two sides to the equation converge to equal the same which always only occurs when the digit sum is 18. The digit sum locks into being at 18 at that point. And then subsequently once the digits are 18 they converge on to 6174. I would start by working out why digits on each side of the equation converge to summing to 18 on both sides of the equation and never being equal at any other value in this process. It reminds me of https://math.stackexchange.com/questions/99725/every-integer-is-congruent-to-the-sum-of-its-digits-mod-9 https://math.stackexchange.com/questions/99725/every-integer... Now the next thing I would do is ask why does every number with digits that sum to 18 eventually end up at 6174. 4 digit numbers with digits that sum to 18 is a very limited set so it should be easy to figure out the combinations and why they all reach 6174. Put those two together and you'd have an answer. (I'm thinking about it now but it really doesn't seem too hard).
- vanderZwan 3y ago
  The follow-up article appears to give some hints though: > A number of readers emailed to say they had discovered that repeatedly adding up the digits of any of the kernels of Kaprekar's operation always equalled 9 (...) Professor Nishiyama has provided an explanation why this happens: it is because the result of performing Kaprekar's operation on any number is a multiple of 9. https://plus.maths.org/content/pluschat-15 https://plus.maths.org/content/pluschat-15 This is starting to look very similar to "if you repeatedly add all the digits of an integer represented in base 10 representation until you have a single digit, and the result is 3, 6 or 9, then it is divisible by 3". I forgot the exact explanation for that one, but IIRC has to do with the implicit calculation that is embedded in base 10 positional notation, other bases have a different number you can quickly verify the divisibility of this way. So maybe that (the "implicit calculation in base 10 representation" thing) is one part of the explanation. I mean whatever it is, it feels like a mix of all these operations imposing constraints upon each other and interacting with the recursive feedback loop to result in the convergence as an emergent property.
  - Arnavion 3y ago
    >This is starting to look very similar to "if you repeatedly add all the digits of an integer represented in base 10 representation until you have a single digit, and the result is 3, 6 or 9, then it is divisible by 3". I forgot the exact explanation for that one, but IIRC has to do with the implicit calculation that is embedded in base 10 positional notation Yes. The number "xyz" is 100x + 10y + z. Each power of 10 can be split into 1 plus a multiple of 9, ie (x + y + z) + (99x + 9y). The second group where all the components are a multiple of 9 is of course divisible by 9 and by 3. The first group is the sum of the digits. So if the sum of the digits is divisible by 3, then the original number was divisible by 3. If the sum of the digits is also divisible by 9, then the original number was also divisible by 9. If the sum of the digits is not divisible by 3, the original number was not divisible by 3 either. This generalizes to any number of digits, and to any base N for testing divisibility by N-1 or factors of N-1.

t-3 3y ago

Check out https://en.wikipedia.org/wiki/Kaprekar%27s_routine https://en.wikipedia.org/wiki/Kaprekar%27s_routine At a glance, there seem to be some patterns, like how for those bases with a 2-digit Kaprekar number the sum of the digits is base-1. There must be some number theory explanation for it.

bmacho 3y ago

The graph is small. Similar graphs (e.g. this algorithm for 2 or 5 digit, or 3-4 digit numbers for base 9 or 11) have multiple cycles: https://en.wikipedia.org/wiki/Kaprekar's_routine#Kaprekar's_constants_and_cycles_of_the_Kaprekar_mapping_for_specific_base_b https://en.wikipedia.org/wiki/Kaprekar's_routine#Kaprekar's_...

penteract 3y ago

One thing that makes it less surprising is that there are lots of numbers which map to the same result - for example all permutation of a bag of digits. I checked, and there are only 55 distinct results (54 excluding 0000) from applying the process to all 4-digit numbers, which leaves less space for lots of cycles.

underlipton 3y ago

Does it work for different base systems? As someone who failed high school calculus (i.e., a moron), I would imagine that it's an artifact of base-10.