If 6 was 9, or 100 was 64, or M was N

Question

         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}

Answer 1

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.

Answer 2

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.

Answer 3

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.