18
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This has been (renamed and) refocused on boboquack’s unforeseen solution, which deserves its own puzzle.

Right when you thought the good old days of New Mathematics were safely dusted, along comes an unending lesson on similarities of numbers in different bases. For this  lesson  puzzle, 1008 means the digits 100 in base 8, which equals the familiar 64 (in base 10), which itself is represented as 6410.

   Just when is 100 the same as 64?   When isn’t it?!   Just look:

          1008    =  6410 ?  
          10010  =  6416 ?  
          10016  =  6442 ?     Could this go on forever?

   That used 6 different digits in all — 0, 1, 2, 4, 6, 8 — and may be summarißed as...

           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.

   And no, this pattern does not continue forever, not even past L Z ( 6442 ).

But wait,  other patterns can indeed go on forever.   And with fewer different digits.

         M P   =  N Q
         M Q  =  N R
         M R  =  N S
         M S   =  N T
           .
           .
           .

What pattern of M, N, P, Q, R, S, T,  .  .  .   uses the fewest different digits?

How can that be generalized to other numbers of different digits?

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


To quote professor Tom Lehrer from New MathYouTube, “Hooray for New Math!”

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  • 1
    $\begingroup$ Is "summarißed" a typo or intentional? $\endgroup$ – Ian MacDonald May 14 '17 at 22:06
  • 1
    $\begingroup$ "Summarißed" is just an attempt to spell both/neither UKish and/nor USAish ("ß" = "s/z"; someday unicodologists will discover a ligature for "o/ou" as well) $\endgroup$ – humn May 14 '17 at 22:08
  • 6
    $\begingroup$ @humn I fear this puzzle is so simple, so very simple, that only a child can do it. $\endgroup$ – Gareth McCaughan May 14 '17 at 22:23
  • 1
    $\begingroup$ (To anyone feeling like I'm insulting @humn for making a bad puzzle or them for not solving it instantly: please follow that YouTube link.) $\endgroup$ – Gareth McCaughan May 14 '17 at 22:25
  • $\begingroup$ The originally intended puzzle will be re-posed in a way that better distinguishes the originally intended solution $\endgroup$ – humn May 15 '17 at 7:47
11
+50
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A little touch-up to the previous answer gives a proper solution:

I doubt you can do it with just one digit, but feel free to prove me wrong!
\begin{array}{lcl}100_{10}&=&10_{100}\\100_{100}&=&10_{10000}\\100_{10000}&=&10_{100000000}\\100_{100000000}&=&10_{10000000000000000}\\&\vdots&\end{array}
\begin{array}{lcl}200_{10}&=&20_{100}\\200_{100}&=&20_{10000}\\200_{10000}&=&20_{100000000}\\200_{100000000}&=&20_{10000000000000000}\\&\vdots&\end{array}
\begin{array}{lcl}203_{10}&=&23_{100}\\203_{100}&=&23_{10000}\\203_{10000}&=&23_{100000000}\\203_{100000000}&=&23_{10000000000000000}\\&\vdots&\end{array}
Note: the pattern really begins here: \begin{array}{lcl}40203_{10}&=&423_{100}\\40203_{100}&=&423_{10000}\\40203_{10000}&=&423_{100000000}\\40203_{100000000}&=&423_{10000000000000000}\\&\vdots&\end{array}
\begin{array}{lcl}5040203_{10}&=&5423_{100}\\5040203_{100}&=&5423_{10000}\\5040203_{10000}&=&5423_{100000000}\\5040203_{100000000}&=&5423_{10000000000000000}\\&\vdots&\end{array}
\begin{array}{lcl}605040203_{10}&=&65423_{100}\\605040203_{100}&=&65423_{10000}\\605040203_{10000}&=&65423_{100000000}\\605040203_{100000000}&=&65423_{10000000000000000}\\&\vdots&\end{array}
\begin{array}{c}\vdots\end{array}

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  • 2
    $\begingroup$ And you can even make an infinite chain too! $\endgroup$ – ffao May 15 '17 at 2:14
  • 3
    $\begingroup$ (Puzzling should have a badge for solutions that get puzzles refitted to them.) $\endgroup$ – humn May 15 '17 at 7:56

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