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[Second part and bounty challenge appended in May 2020]


          ALL ANIMALS ARE EQUAL
   BUT SOME ANIMALS ARE MORE EQUAL THAN OTHERS


         — from Animal Farm by George Orwell

Original puzzle from March 2019:

A contrived simple equivalence rule A applies neatly to numbers 0 through 99. (Application of this equivalence to further numbers is, well, equivocal.) All equivalences headed by numbers 0 through 19 are listed below, accounting for almost all other eligible numbers as well, where ‘=’ means “is equivalent to.” (Each number is reflexively equivalent to itself.)

$\require{begingroup}\begingroup \def\b #1{{ \bf\phantom{39}\llap{#1} }} \def\no {{ \textsf{no others} }} \def\= {{ \tiny \raise.3ex{\: = ~~} }} \small\begin{array}{llll} \textsf{Equivalence rule A:} ~~~ & \b{0} \= \no && \b{10}\= \no \\ & \b{1} \= \no && \b{11}\= 29 \= 31 \= 49 \= 51 \= 69 \= 71 \= 89 \= 91 \\ & \b{2} \= \no && \b{12}\= 28 \= 32 \= 48 \= 52 \= 68 \= 72 \= 88 \= 92 \\ & \b{3} \= \no && \b{13}\= 27 \= 33 \= 47 \= 53 \= 67 \= 73 \= 87 \= 93 \\ & \b{4} \= \no && \b{14}\= 26 \= 34 \= 46 \= 54 \= 66 \= 74 \= 86 \= 94 \\ & \b{5} \= \no && \b{15}\= 25 \= 35 \= 45 \= 55 \= 65 \= 75 \= 85 \= 95 \\ & \b{6} \= \no && \b{16}\= 24 \= 36 \= 44 \= 56 \= 64 \= 76 \= 84 \= 96 \\ & \b{7} \= \no && \b{17}\= 23 \= 37 \= 43 \= 57 \= 63 \= 77 \= 83 \= 97 \\ & \b{8} \= \no && \b{18}\= 22 \= 38 \= 42 \= 58 \= 62 \= 78 \= 82 \= 98 \\ & \b{9} \= \no && \b{19}\= 21 \= 39 \= 41 \= 59 \= 61 \= 79 \= 81 \= 99 \\ & && \b{20}\= \underline{~~~~~~~~} \dots \bf \, ? \\[-.5ex] & &~~~~& \phantom{2}\vdots \end{array}$

  1. What, if any, are the equivalences of 20 by rule A?
    Please use and explain the simplest possible rule (which can be described in 11 words or less and is not purely mathematical) that accounts for every equivalence from 0 to 99.


Note: The equivalences listed above and below are complete and not equivalent to any other numbers.


Related puzzle, added partly to serve as a hint in May 2020:

A contrived simple equivalence rule B applies neatly to numbers 1 through 3999. (Application of this equivalence to further numbers is again equivocal.) All equivalences headed by numbers 1 through 39 are listed below, accounting for many other eligible numbers as well, with some extra spacing to help distinguish groups of consecutive numbers.

$\small\begin{array}{ll} \textsf{Equivalence rule B:} ~~ &\b{ 1} \= 2 \= 3 \\[.1ex] &\b{ 4} \= 5 \= 6 \= 7 \= 8 \\[.3ex] &\b{ 9} \= 10 \= 11 \= 12 \= 13 \\[.5ex] &\b{14} \= 15 \= 16 \= 17 \= 18 ~~ \= ~~ 50 \= 51 \= 52 \= 53 \\[.7ex] &\b{19} \= 20 \= 21 \= 22 \= 23 \kern1.1em \= ~~ 100 \= 101 \= 102 \= 103 \\[.9ex] &\b{24} \= 25 \= 26 \= 27 \= 28 \kern1.9em \= ~~ 40 \= 41 \= 42 \= 43 \\ &\b{} \= 59 \= 60 \= 61 \= 62 \= 63 \kern0.6em \= ~~ 104 \= 105 \= 106 \= 107 \= 108 \\ &\b{} \= 500 \= 501 \= 502 \= 503 \\[.9ex] &\b{29} \= 30 \= 31 \= 32 \= 33 \kern3.8em \= ~~ 90 \= 91 \= 92 \= 93 \\ &\b{} \= 109 \= 110 \= 111 \= 112 \= 113 ~~ \= ~~ 1000 \= 1001 \= 1002 \= 1003 \\[.9ex] &\b{34} \= 35 \= 36 \= 37 \= 38 \!\:~~~ \= ~~\!\: 49 ~~ \= ~~ 69 \= 70 \= 71 \= 72 \= 73 \\ &\b{} \= 94 \= 95 \= 96 \= 97 \= 98 \!\:~~~~ \= ~~ 114 \= 115 \= 116 \= 117 \= 118 \\ &\b{} \= 150 \= 151 \= 152 \= 153 ~~ \= ~~ 509 \= 510 \= 511 \= 512 \= 513 \\ &\b{} \= 1004 \= 1005 \= 1006 \= 1007 \= 1008 \\[.8ex] &\b{39} \= ~~ 99 ~~ \= ~~ 119 \= 120 \= 121 \= 122 \= 123 ~~ \= ~~ 200 \= 201 \= 202 \= 203 \\ &\b{} \= 1009 \= 1010 \= 1011 \= 1012 \= 1013 \\[.6ex] &\b{44} \= \underline{~~~~~~~~} \dots \bf \, ? \\[-.5ex] & \phantom{4}\vdots \end{array}\endgroup$

  1. What, if any, are the equivalences of 44 by rule B?

  2. Which number /numbers is /are the  l e a s t  equivalent (equivalent to the fewest others) by rule B?

Bounty challenge toughie, not required for a  ✓ correct answer but nonetheless:

  1. Which numbers are the  m o s t  equivalent (equivalent to the most others) by rule B?
    (Anything close deserves votes of approval. A reasoned attempt deserves a bounty amount relative to thoroughness and correctness.)
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    $\begingroup$ Hurray, a humn puzzle! It's been a while. $\endgroup$ Mar 24 '19 at 19:06
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    $\begingroup$ Is there a way to "watch" a question so that I'm notified of new or accepted answers? I've already starred it. $\endgroup$
    – MooseBoys
    Mar 27 '19 at 1:04
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    $\begingroup$ To be clear, is the relation really only meaningful for numbers 0 thru 99, or are you just saying all numbers outside that range would be "no others"? $\endgroup$
    – MooseBoys
    Mar 27 '19 at 1:07
  • $\begingroup$ Thank you, @MooseBoys. Yes the relation only has relevance to numbers 0 through 99. "No others" would indeed be a great catch-all for other numbers. $\endgroup$
    – humn
    Mar 27 '19 at 12:17
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Answer:

20 = no others

Reason: (humn has told me that this is wrong but it's my favorite guess of mine)

Because you gave us a list of equivalences which are more equal than others. So we can assume the remaining numbers are less equal and therefore only equal to themselves.


Other guesses:

Xilpex's rule applies if no digits are zero. If any digit is zero (2 can be written as 02) then there are no equivalents

Because the rules are contrived so I can simply invent whatever I want for the rules that aren't given to me.

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  • $\begingroup$ Correct answer, @ferret! But the reasoning is more complicated than necessary. $\endgroup$
    – humn
    Mar 25 '19 at 2:02
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    $\begingroup$ @humn edited with a new "lateral thinking" attempt $\endgroup$
    – ferret
    Mar 25 '19 at 3:04
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    $\begingroup$ You're on the way, @ferret, and gave me an idea for another puzzle. Still missing the essential ingredient. $\endgroup$
    – humn
    Mar 25 '19 at 4:06
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    $\begingroup$ @humn is it because they are rot13 pbagevirq? $\endgroup$
    – ferret
    Mar 25 '19 at 5:31
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    $\begingroup$ Thank you for playing along, @ferret. Pleasure to have met you. $\endgroup$
    – humn
    Mar 25 '19 at 17:06
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20 = no others. same as 0 to 10. because it breaks the following pattern. The green cells are the numbers given and the colored cells is the sum their digits above them. Numbers 0 - 10 break the pattern of the sums as well. Hence the "no others" as they don't follow the pattern of the sums as others
enter image description here
i know there is no 100, 101, i used excel for this, regardless it still doesn't follow the pattern of the sums either way

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  • $\begingroup$ Congratulations on the correct answer. And what a wonderfully visual and consistent explanation. Yet the intended explanation is simpler and not nearly as gorgeous. $\endgroup$
    – humn
    Apr 7 '19 at 3:21
  • $\begingroup$ @humn well, the non mathematical explanation would be that numbers with leading or trailing zeroes have no equivalence. 00,01,02,03,04,05,06,07,08,09,10,20 $\endgroup$
    – Mel
    Apr 7 '19 at 9:38
  • $\begingroup$ Your non mathematical 0s approach does fit the pattern, @Mel, but doesn't explain the equivalences as well as your excel solution. $\endgroup$
    – humn
    Apr 7 '19 at 11:33
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  0 = no others      ­ 10 = no others      ­ 20 = no others
  1 = no others      ­ 1 1 = 2 9 = 3 1 = 4 9 = 5 1 = 69 = 71 = 89 = 91
  2 = no others      ­ 1 2 = 2 8 = 3 2 = 4 8 = 5 2 = 68 = 72 = 88 = 92
  3 = no others      ­ 1 3 = 2 7 = 3 3 = 4 7 = 5 3 = 67 = 73 = 87 = 93
  4 = no others      ­ 1 4 = 2 6 = 3 4 = 4 6 = 5 4 = 66 = 74 = 86 = 94
  5 = no others      ­ 1 5 = 2 5 = 3 5 = 4 5 = 5 5 = 65 = 75 = 85 = 95
  6 = no others      ­ 1 6 = 2 4 = 3 6 = 4 4 = 5 6 = 64 = 76 = 84 = 96
  7 = no others      ­ 1 7 = 2 3 = 3 7 = 4 3 = 5 7 = 63 = 77 = 83 = 97
  8 = no others      ­ 1 8 = 2 2 = 3 8 = 4 2 = 5 8 = 62 = 78 = 82 = 98
  9 = no others      ­ 1 9 = 2 1 = 3 9 = 4 1 = 5 9 = 61 = 79 = 81 = 99
Delete the tens digit, like follow:

  0 = no others      ­ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
  1 = no others      ­ 1 = 9 = 1 = 9 = 1 = 9 = 1 = 9 = 1
  2 = no others      ­ 2 = 8 = 2 = 8 = 2 = 8 = 2 = 8 = 2
  3 = no others      ­ 3 = 7 = 3 = 7 = 3 = 7 = 3 = 7 = 3
  4 = no others      ­ 4 = 6 = 4 = 6 = 4 = 6 = 4 = 6 = 4
  5 = no others      ­ 5 = 5 = 5 = 5 = 5 = 5 = 5 = 5 = 5
  6 = no others      ­ 6 = 4 = 6 = 4 = 6 = 4 = 6 = 4 = 6
  7 = no others      ­ 7 = 3 = 7 = 3 = 7 = 3 = 7 = 3 = 7
  8 = no others      ­ 8 = 2 = 8 = 2 = 8 = 2 = 8 = 2 = 8
  9 = no others      ­ 9 = 1 = 9 = 1 = 9 = 1 = 9 = 1 = 9
So there is no rules to 0,

20 = no others

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    $\begingroup$ Keep going, @user58107! It's simpler than that. $\endgroup$
    – humn
    Mar 25 '19 at 6:42
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    $\begingroup$ look the column, my English very poor, can't explain clarification. $\endgroup$
    – user58107
    Mar 25 '19 at 6:46
  • $\begingroup$ Oh, oh oh oh, @user58107, this puzzle relies on English. (Big give-away.) Thank you for hitching the ride. $\endgroup$
    – humn
    Mar 25 '19 at 6:52
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    $\begingroup$ @humn Maybe a language tag then? $\endgroup$
    – Rubio
    Apr 6 '19 at 4:07
  • $\begingroup$ Right, @Rubio, language tag added. I was trying to not give away that aspect but did in the comment above and the time is ripe anyway. $\endgroup$
    – humn
    Apr 7 '19 at 3:11
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My first thought was

The equivalence classes are based on the distance of a number to the closest multiple of 20: $$|11 - 20| = 9, \quad |29 - 20| = 9,\quad |31 - 40| = 9,\quad \ldots$$

However, that did not explain 0 through 10. I could add a 'except for 1 through 10' to my rule, but that wasn't very satisfying.

The second thing I came up with was:

For a number $n$ made of two digits $a$ and $b$, we have $n = a\cdot 10 + b$. If we say those are equivalent to $m=a\cdot 10 - b$, they are recursively equivalent to a lot of numbers. For example, $$97 = 9\cdot 10 + 7$$ $$9\cdot 10 - 7 = 83 = 8\cdot 10 + 3$$ $$8\cdot10 - 3 = 77 = 7\cdot 10 + 7$$ And so on: $97 \rightarrow 83 \rightarrow 77 \rightarrow 63 \rightarrow 57 \rightarrow 43 \rightarrow 37 \rightarrow \ldots$. This shows that numbers with a zero as second digit do not have any other equivalent numbers, as this process would not lead to any new numbers: $$20 = 2\cdot 10 + 0$$ $$2\cdot10 - 0 = 20$$ $20 \rightarrow 20 \rightarrow \ldots$. The only way I managed to explain $1,\ldots,9$ here is to explain you simply can't apply this process to those numbers, as they do not have two digits.

This doesn't work if you accept that 09 is a perfectly fine way of writing 9. So I'm still not very satisfied with this solution.

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  • $\begingroup$ M-ou-se, the indended explanation is hidden inside your second approach (about multiples of 10). That is the closest any has yet gotten! Still, the intended solution is simpler and not quite as mathematical. $\endgroup$
    – humn
    Apr 7 '19 at 11:42
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Some musings below:

The first thing that strikes us as peculiar about rule B is:

That it only applies for numbers from 1 to 3999

Aha! Perhaps the rule is about

Roman numerals?

In particular,

It looks like the number of Is in a particular number's representation is inconsequential.

So, perhaps these sets represent:

The numbers from 0 onwards in order?

To find such a bijection,

Consider I→0, V→1, X→2, L→3, C→4, D→5 and M→6 and add up the letters within a particular number's representation.

This does account for all the equivalences in the rule. However,

Some numbers are missing, such as 54 through 58 which would have a total of (ignoring any Is) LV→3+1=4 so should fit under 19's header (19=XIX→2+0+2=4).

Under this (flawed) approach, the answer to question 2 would then be:

44=XLIV→2+3+0+1=6 so it should fit under 29's equivalence class. Under this interpretation, a full (I think) list of numbers in this class would be 29 through 33 (XXX), 44 through 48 (XLV), 64 through 68 (LXV), 89 through 93 (XC), 109 through 113 (CX), 504 through 508 (DV) and 1000 through 1003 (M).

If a similar approach were to be applied to the original question,

We might be looking for a number system going from 0 to 99 with numbers represented by their distance to the nearest multiple of 20, such that multiples of 20 all have the same non-zero "value", but I can't think of an applicable system off the top of my head.


Further digging suggests:

The missing values are those with two "multiple of 5 counters", i.e. two of those of VLD. (Perhaps those with more are missing, but we would need to get up to a sum of 9 to confirm.)

Using this reasoning, perhaps:

Those values are in some separate category for some yet-to-be understood reasoning.

In which case the answer to question 2 would be:

44 would be equal to 44 through 48 (XLV), 64 to 68 (LXV) and 504 through 508 (DV).

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  • $\begingroup$ @humn How curious! I shall experiment some more in that case. $\endgroup$
    – boboquack
    May 22 '20 at 8:23
  • $\begingroup$ @humn Made some more observations of questionable usefulness... but it's still sorely in need of a Grand Unified Theory. $\endgroup$
    – boboquack
    May 22 '20 at 8:39
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20 would be:

20 = 20 = 40 = 40 = 60 = 60 = 80 = 80 = 100

Explanation:

The rule (vertically) is: Line 1 + 1, then Line 2 - 1, and so on.

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    $\begingroup$ Thank you for taking the bait, Xilpex. Not quite the solution, though. For instance, it doesn't explain the entry for 10. $\endgroup$
    – humn
    Mar 24 '19 at 18:23
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    $\begingroup$ @humn Ok. I'll see if there is any other answer... :D $\endgroup$
    – xilpex
    Mar 24 '19 at 18:25
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    $\begingroup$ Plus there is no $100$. $\endgroup$ Mar 24 '19 at 18:36

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