S2T2

Jim Randell 4:43 pm on 23 October 2020 Permalink Reply
Tags: by: Howard Williams ( 33 )   

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  Jim Randell 4:59 pm on 23 October 2020 Permalink | Reply

  See also: Enigma 1036, Enigma 1161, Teaser 2565.

  If we have a number, where the leading digit is a and the remaining k digits are bcdefg… = r, then we have:

  r.10 + a = (25/43)(a.10^k + r)

  The following Python program runs in 45ms.

  Run: [ @replit ]

  from enigma import (irange, div, int2base, printf)
  
  # consider k digit numbers
  for k in irange(1, 9):
    m = 10 ** k
    # consider possible leading digit
    for a in irange(1, 9):
      r = div(a * (25 * m - 43), 405)
      if r is None or not (r < m): continue
      # output solution
      printf("{a}{r} -> {r}{a}", r=int2base(r, width=k))
  

  Solution: The correct number is: 530864197.

  So the number entered is: 308641975.

  >>> fraction(308641975, 530864197)
  (25, 43)
  

  ---

  For this particular puzzle we can do some analysis can reduce the solution space further. (From the 9×9 = 81 cases to consider in the general case).

  In the equation:

  43(r.10 + a) = 25(a.10^k + r)

  we see that that each side is divisible by 5, so a must be 5 (as it cannot be 0).

  Which leaves:

  r = (25.10^k − 43) / 81

  Which we can then check with the 9 possible values for k manually or with an even shorter program:

  from enigma import (irange, div, int2base, printf)
  
  for k in irange(1, 9):
    r = div(25 * 10 ** k - 43, 81)
    if r is not None:
      printf("5{r} -> {r}5", r=int2base(r, width=k))
  

  Or, for a complete manual solution:

  Numbers of the form: (25.10^k − 43) / 9, look like this:

  k=1: 23
  k=2: 273
  k=3: 2773
  k=4: 27773
  …

  To divide by 9 again we need the number to have a digital root of 9, and the only one in the required range is:

  k=8: 277777773

  Dividing this by 9 gives:

  r = 277777773 / 9 = 30864197

  Hence the correct number is 530864197, and the incorrect number 308641975.

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    hans droog 10:01 am on 6 November 2020 Permalink | Reply

    Hi Jim. would be obliged if you could explain formula in teaser 3031. Many thanks, Hans Droog

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      Jim Randell 10:10 am on 6 November 2020 Permalink | Reply

      @hans:

      As an example, if we had an 7 digit number abcdefg which was entered incorrectly on the calculator as bcdefga, then we would have:

      bcdefga = (25/43)(abcdefg)

      If we represent the 6 digit number bcdefg as r we can write this expression as:

      r.10 + a = (25/43)(a.10^6 + r)

      The expression I give in my solution is the general case when r is a k digit number.

      (“.” is multiplication. “^” is exponentiation).

      Is that clear?

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  Frits 1:42 pm on 24 October 2020 Permalink | Reply

  Similar.

  print(["5"+str(lastpart // 81) for k in range(2, 10) 
        if (lastpart := (25 * 10**k - 43)) % 81 == 0][0])
  

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  Hugh Casement 4:15 pm on 7 November 2020 Permalink | Reply

  Hans may have been put off by Jim’s use of a decimal point to mean multiplication.
  The international convention is a raised dot.

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    Jim Randell 10:13 am on 8 November 2020 Permalink | Reply

    If I were handwriting a decimal number I would use a “middle dot” for the decimal point, and a dot on the line to denote multiplication. Of course when typing the “.” (full stop) symbol has to do for both.

    Fortunately we don’t often have to deal with numbers with decimal points in puzzles.

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  hans droog 9:01 am on 8 November 2020 Permalink | Reply

  thats right Hugh

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