14
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         Now if a 6,
         Turned out to be 9,
         I don’ t mind,
         I don’ t mind   – Jimi Hendrix in If 6 was 9

Cool. Suppose a 100 turned out to be 64.

         I don’ t care,
         I don’ t care   – Jimi Hendrix in If 6 was 9

But, but Jimi, what if your computer program is bugged out and can’t tell if its bad self is coming (from octal) or going (to hex)?

          1008    =  6410
          10010  =  6416
          10016  =  6442

         Let it be,
         It ain’ t me   – Jimi Hendrix in If 6 was 9

You know, 1008 stands for the digits 100 in base 8, equaling the familiar 64 (in base 10), which itself shows as 6410.

         Dig    – Jimi Hendrix in If 6 was 9

So, that used 6 different digits — 0, 1, 2, 4, 6, 8 — a total of 26 times and bubbles down to...

         K  W  =  L X
         K X   =  L Y
         K Y   =  L Z
         ...where   K = 100,   L = 64,   W = 8,   X = 10,   Y = 16   and   Z = 42.

         I’ m gonna wave my freak flag high! – Jimi Hendrix in If 6 was 9


       Hold tight, hitch-hiker, check out where simpler ingredients can take you.


Just 4 different digits from 0 through 9, taken fewer than 43 times total, can really wig you out.

         M P   =  N Q     ( M > N   and   P < Q < R < S < T )
         M Q  =  N R
         M R  =  N S
         M S   =  N T

     Wild.   What M, N, P, Q, R, S and T could do that with the fewest digits total?

         Even a 43-digits-total solution or two would be worth posting.

         Wave on,
         Wave on . . .
         . . .’ Cause everybody knows what I’ m talking about
          – Jimi Hendrix in If 6 was 9 (vocal track) dailymotion

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  • $\begingroup$ Computers are in play for calculations, of course, just not for combinatorial searches. This puzzle reincarnates the original intention of New Mathematics forever. $\endgroup$ – humn May 19 '17 at 15:32
  • $\begingroup$ I feel very dumb, but is 100 in base 16 really the same as 64 in base 42? $\endgroup$ – mfloren Jun 12 '17 at 5:31
  • $\begingroup$ $100_{16} = 256_{10}= 6\times 42 + 4$, this time $\endgroup$ – humn Jun 12 '17 at 5:36
  • $\begingroup$ :) Thank you, completely my mistake: I was using mods instead of bases... Darn you, new math! $\endgroup$ – mfloren Jun 12 '17 at 5:38
  • 1
    $\begingroup$ It's so simple, so very simple, that only a child can do it $\endgroup$ – humn Jun 12 '17 at 5:42
3
+250
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Well, I have a working set, it seems:

1011 (base 1) = 11 (base 2)
1011 (base 2) = 11 (base 10)
1011 (base 10) = 11 (base 1010)
1011 (base 1010) = 11 (base 1030302010)

using 4 digits (0,1,2 and 3) and the total number of times those numbers used is - 49 times (well, it was above the 43 limit...) and

assuming fractional bases are valid, here is another one:

22 (base 2.25) = 11 (base 5.5)
22 (base 5.5) = 11 (base 12)
22 (base 12) = 11 (base 25)
22 (base 25) = 11 (base 51 )

using 3 digits (1, 2 and 5) and the total number of times those numbers used is 33.

Here is another one:

111 (base 2) = 21 (base 3)
111 (base 3) = 21 (base 6)
111 (base 6) = 21 (base 21)
111 (base 21) = 21 (base 231)

using 4 digits (1, 2, 3 and 6) and the total number of times those numbers used is 32. Hope this satisfies @humn's requirements in all manners.

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  • 4
    $\begingroup$ "base 1" has (by definition) only one possible digit value, so your first line doesn't work. $\endgroup$ – Rubio Aug 3 '17 at 16:53
  • $\begingroup$ @Rubio what is that possible value Rubio (of base 1 system)? $\endgroup$ – Mea Culpa Nay Aug 4 '17 at 7:45
  • 1
    $\begingroup$ I've argued for a long that 0 should be a valid digit in base 1. Would be consistent with all the other bases, including irrational, imaginary and complex. A 0 at each $n$th place merely adds 0 times $1^n$, while each 1 at each $m$th place adds 1 times $1^m$ as usual. $\endgroup$ – humn Aug 4 '17 at 9:06
  • 1
    $\begingroup$ I really to love this solution! (Even though uses more digits than called for.) $\endgroup$ – humn Aug 4 '17 at 9:10
  • 1
    $\begingroup$ "Base 1" is in effect a tally system, with 1 usually chosen as the digit to use. 1 is "1", 2 is "11", 3 is "111", and so on. Cnsider the usual meaning of Base N: each position in a number, right to left, is an increasing power of N, with N distinct digits in use; base 10 has digits 0..9, and positions (right to left) of 10⁰ (ones), 10¹ (tens), 10² (hundreds), and so on. Base 1 should thus have just a single digit, and position values 1⁰, 1¹, 1², … — all of which equal 1, so you end up representing a value V as a sequence of V 1's. (0 skips a digit position; no need/use for that in Base 1.) $\endgroup$ – Rubio Aug 5 '17 at 7:30
3
+100
$\begingroup$

After being told I misread the question, again, and that 4 digits was a requirement, not a max, I'm modifying my solution as explained below. The new new solution with 73 digits is:

M = 203, N = 23 P, Q, R, S and T are all the square of the previous value. Something as simple as 10 has all its powers use the same 2 digits, and fit within the parameters, resulting in 4 total distinct digits. P = 10 = 1E1, Q=100=1E2, R=1E4, S=1E8, and T=1E16. That should yield 49 total zeros, and 8 ones, 8 twos and 8 threes, meaning 73 total digits.

Below was what I wrote when I thought that the solution was based on a max of 4 digits:

After being told I totally misread the problem, I have totally new solution. It uses 73 total digits (only 2 distinct digits). I think it's possible to narrow that down, but it probably involves something interesting with storing P-T in non-base 10, or using symbols that represent digits higher than 9.

M = 100, N = 10 P, Q, R, S and T are all the square of the previous value. Something as simple as 10 has all its powers use less then 4 digits, and fit within the parameters, P = 10 = 1E1, Q=100=1E2, R=1E4, S=1E8, and T=1E16. That should yield 57 total zeros, and 16 ones, meaning 73 total digits.

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  • $\begingroup$ Looks like you solved a different puzzle! Not bad. (Might be worth posting it separately. I'd like to see it.) But your approach looks promising for this puzzle too. The way your solution lays out according to this puzzle matches my best hand-figured 32 total digits (most values get counted twice or more), and the first line of $\small M_{_P}=N_{_Q}$ comes out $\small 55_{35} = 54_{40}$, which works out to an untrue $\small 180 \ne 204$ according the rules here. $\endgroup$ – humn Aug 4 '17 at 9:31
  • $\begingroup$ Sorry, totally misunderstood the problem. Putting a new solution in. $\endgroup$ – Jesse Aug 4 '17 at 22:17
  • $\begingroup$ Well, @Jesse, the proposed solution mustuse 4 different digits. Your solution uses only 2!. Please check. $\endgroup$ – Mea Culpa Nay Aug 5 '17 at 1:00
  • $\begingroup$ And now you solved the original version of this puzzle, New Mathematics forever, Jesse! That's actually a more clever solution than for this one, which is why this version asks for 4 different digits and has a qualitatively different solution. $\endgroup$ – humn Aug 5 '17 at 1:25
  • $\begingroup$ @humn a solution that requires 4 different digits (thought that was a max), is trivially generated from my solution. Multiplying M and N by 2 and adding 3 to each will keep the same bases, the same total digits and will get us to 4 digits. Will modify solution to match. $\endgroup$ – Jesse Aug 5 '17 at 4:03
2
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M = 121, N = 100, P = 10, Q = 11, R = 12, S = 13, T = 14

I know this solution uses 5 digits (a total of 40 digits), but that was a quick manual find.

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