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

     Wild. &hairsp; 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 . . .
         . . .’&hairsp;Cause everybody knows what I’&hairsp;m talking about
        &hairsp; – Jimi Hendrix in If 6 was 9 (vocal track) dailymotion

humn
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3 Answers3

4

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.

Mea Culpa Nay
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    "base 1" has (by definition) only one possible digit value, so your first line doesn't work. – Rubio Aug 03 '17 at 16:53
  • @Rubio what is that possible value Rubio (of base 1 system)? – Mea Culpa Nay Aug 04 '17 at 07:45
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    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. – humn Aug 04 '17 at 09:06
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    I really to love this solution! (Even though uses more digits than called for.) – humn Aug 04 '17 at 09:10
  • Thanks @humn, a nice question (but I guess solution is possible only through brute force / trial and error means only) ! – Mea Culpa Nay Aug 04 '17 at 10:12
  • True that my solutions weren't purely deductive. They were pretty much intuitive and involved almost no trial and error. Plus a little luck, admittedly, which comes with intuition, upon which I typically rely when doing mathematics. In any case a bounty is coming for this solution because it "is neat and pretty, until a phenomenal climactic use of base 1030302010 !" I'm guessing that you used some intuition and luck, too Mea Culpa Nay. – humn Aug 05 '17 at 01:44
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    "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.) – Rubio Aug 05 '17 at 07:30
  • Nice comparison to Roman numerals, @Rubio! – humn Aug 05 '17 at 14:21
  • I just reviewed my two 43-total-digits solutions, Mea Culpa Nay, and now actually claim they were honestly derived from a familiarity with powers of 5. One of them, which I might post as a wiki, uses a fractional base that you might enjoy as much as I enjoy your use of base 1030302010. (My alas-not-optimal 32-total-digits solution, though, remains a matter of intuition and luck, but more power to anyone's finding or beating it without a computer.) [Sorry about the deleted comments on the way to this one, if you were notified of them.] – humn Aug 05 '17 at 14:30
  • Well, @Rubio, a base 10 system has 10 digits (0 through 9), a base 8 system has 8 digits (0 through 7) ...and in the same manner..A base 2 system has two digits 0 and 1 ...And hence a base 1 system should have only one digit and it should be '0'...:-), though it does not make sense...As such unary systems...are complemented by something which represents 'absence of any' or 'presence of nothing' by '0'. Hence 0 and 1 should form the digits in an unary system. Anyways, these are my own views which may...be inline or out of line with others. – Mea Culpa Nay Aug 06 '17 at 06:52
  • @humn Hi, is a bounty announced on this or not yet ? I have a solution ready with 4 digits and less than or equal to 40 times the used numbers counted. :-) – Mea Culpa Nay Aug 11 '17 at 07:52
  • @humn, I have added my 2nd solution below my first one. Please check it. – Mea Culpa Nay Aug 18 '17 at 15:37
  • Err...the 2nd one got slipped...through checks...Need to get a different one @humn...Thanks Rubio! – Mea Culpa Nay Aug 19 '17 at 11:55
  • Fractional bases are certainly welcome here, Mea Culpa Nay, wow! (One of my 43-count solutions uses a single fractional base, and you went and found a much smaller solution that uses two.) You have such a refreshing feel for this puzzle. A bounty is on the way, partly as a bribe for you to add brief mentions of how you came up with these solutions. Too bad about 2 vs 3 in the previous solution that included a perfect ${\small 10}_{10} , \small .$ [And sorry about the multiple versions of this comment.] – humn Aug 19 '17 at 16:35
  • @humn, thanks for the bounty ! Well, solutions ... though involved trial and error, but as you told intuition also played a role in this. For example, thinking over a while suggests that the choice of 4 digits (based on which P,Q,R and S can be arrived at) decides M and N. Also, instead of incrementing by one, doubling / doubling and and then subtracting one patterns proved to be helpful while selecting M and N, though the actual/intended solution which is 4 digits based and uses less than 43 times of usage of digits still illudes me :-). – Mea Culpa Nay Aug 22 '17 at 03:53
  • Thank you for more than one reason to bounty, Mea Culpa Nay. Care to add the bulk of your latest comment (or something like it) to the solution post itself? Perhaps even include what each solution looked like midway. These kinds of notes could just be added recklessly to the end of the post as something like "Addendum. Unedited notes and worksheets." Not to reap 1000s of points from this puzzle's limited audience but to broaden the imaginations of current and future members of that audience. Your solutions are spectacularly mystifying examples of lateral-thinking! – humn Aug 22 '17 at 12:43
  • Today's solution gets the $\color{#0c0}{\large\raise-.1ex\checkmark}\kern-.3em$, Mea Culpa Nay, right on! It's not the absolutely minimal solution, which was found with a computer, but it is exquisitely neat and, I hope, the minimum that anyone would find by hand. By the way, I'd still like to see added to your answer hints at how you came up with the first two fantastic solutions. – humn Sep 11 '17 at 18:11
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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.

2012rcampion
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Jesse
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  • 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. – humn Aug 04 '17 at 09:31
  • Sorry, totally misunderstood the problem. Putting a new solution in. – Jesse Aug 04 '17 at 22:17
  • Well, @Jesse, the proposed solution mustuse 4 different digits. Your solution uses only 2!. Please check. – Mea Culpa Nay Aug 05 '17 at 01:00
  • 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. – humn Aug 05 '17 at 01:25
  • @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. – Jesse Aug 05 '17 at 04:03
  • Oops, the comment above should be also directed @MeaCulpaNay – Jesse Aug 05 '17 at 05:51
  • When calculating the number of digits I think you have to take into account the fact that M and N occur four times and Q R and S occur twice. – 2012rcampion Aug 05 '17 at 13:19
  • Beautiful 4-different-digit adaptation, which can continue infinitely! And nicely derived. (I really enjoy how this puzzle has so many surprises.) Incidentally, my total-digits count for this latest solution comes out 73 (49 0s, 8 1s, 8 2s, 8 3s) but this solution will get the next bounty when the system allows and I log in again. If you don't mind, Jesse, I'd like to sometime edit/add a tabular display to this solution. – humn Aug 05 '17 at 14:13
  • good catch by @2012rcampion on 5th Aug. 2017, though that was what explained by humn in the original problem posted. – Mea Culpa Nay Aug 06 '17 at 17:00
2

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.

Kalaivanan
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