S2T2

Jim Randell 8:04 am on 19 May 2024 Permalink Reply
Tags: by: Oliver Anderson ( 3 )   

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  Jim Randell 8:11 am on 19 May 2024 Permalink | Reply

  See: Puzzle #180, Enigma 653, Enigma 924, Teaser 2565, Enigma 1036, Enigma 455, Enigma 1161 (but note these may reveal the answer to this puzzle).

  Normally a parasitic number [ @wikipedia ] involves multiplying the number by a(n integer) value by moving the final digit to the front.

  This puzzle moves the front digit to the end, so we are looking at the inverse of the process involved in parasitic numbers. So if we find a 2-parasitic number, we can start with it, move the front digit (back) to the end, and it is divided by 2.

  For this Python program I adapted my [[ parasitic() ]] function from Enigma 924 to deal with fractional multipliers, and this allows us to solve both problems mentioned in the puzzle text.

  This program runs in 59ms. (Internal runtime is 167µs).

  Run: [ @replit ]

  from enigma import (irange, inf, div, ndigits, first, printf)
  
  # find numbers such that moving the final digit to the front
  # result in a multiplication of the original number by k/q
  def parasitic(k, q=1, M=inf, base=10):
    assert k < q * base and q < k * base
    b = base * k - q
  
    # consider n digit numbers
    for n in irange(1, M):
      a = base**n * q - k
      m = base**(n - 1)
      # consider non-zero digits d
      for d in irange(1, base - 1):
        x = div(d * a, b)
        if x is not None and m * base > x >= m:
          yield base * x + d
  
  # find numbers such that moving the first digit to the end
  # result in a multiplication of the original number by p/q
  # this is the same as finding a (q/p)-parasitic number
  for (p, q) in [(3, 2), (1, 2)]:
    # find the lowest number
    for x in first(parasitic(q, p), 1):
      n = div(x * q, p)
      # output solution
      printf("[{p}/{q}] {n} * {p}/{q} = {x} [{k} digits]", k=ndigits(n))
  

  Solution: M has 18 digits.

  The smallest number is 210526315789473684.

  210526315789473684 × 1/2 = 105263157894736842

  Note that this is the repetend of 4/19 = 0.(210526315789473684)…

  If you don’t mind leading zeros we can apply the same procedure to the result (which also has 18 digits), to get:

  105263157894736842 × 1/2 = 052631578947368421

  See also: OEIS A337922, OEIS A146088.

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  Frits 3:24 pm on 19 May 2024 Permalink | Reply

    
  # k-digit number n = fp = 2(10p + f) with 19p = (10^(k - 1) - 2)f
  # 19p = a999...999b with a = f - 1 and b = 10 - 2f
  
  for k in range(2, 99):
    for f in range(2, 6):
      # check for a multiple of 19
      if (p19 := f * 10**(k - 1) - 2 * f) % 19: continue
      s19 = str(p19)
      a = str(f - 1)
      b = str(10 - 2 * f)
      
      if s19[1:-1] == "9" * (k - 2) and (s19[0], s19[-1]) == (a, b):
        print("answer:", k)
        break
    else:
      continue
    break   
  

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