S2T2

Jim Randell 1:39 pm on 2 September 2025 Permalink Reply
Tags: by: Danny Roth ( 84 )   

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  Jim Randell 1:41 pm on 2 September 2025 Permalink | Reply

  I often forget about the [[ choose() ]] function in the enigma.py library (see: Tantalizer 482), but this is a good chance to use it. And this gives us a nice compact program.

  The ages must all be different, as the first reversed sum formed must be divisible by A, but none of the others. Hence A is different from all the others. Similarly the second reversed sum is divisible by B, but not C, D, E. Hence B is different from all the others. And so on. So we can use the [[ distinct=1 ]] parameter to the [[ choose() ]] function.

  It wasn’t completely clear if the “2-digit” condition was meant to apply to all the reversed sums, or just the first. I applied it to all of them, although it doesn’t change the answer if this condition is ignored completely.

  This Python program runs in 62ms. (Internal runtime is 326µs).

  from enigma import (irange, choose, nrev, printf)
  
  # check a set of numbers
  def fn(*ns):
    # reverse the sum
    r = nrev(sum(ns))
    # must be a 2 digit number, divisible by just the last number
    return (9 < r < 100 and r % ns[-1] == 0 and all(r % n > 0 for n in ns[:-1]))
  
  # choose 5 single digit ages
  for (E, D, C, B, A) in choose(irange(1, 9), [None, fn, fn, fn, fn], distinct=1):
    printf("A={A} B={B} C={C} D={D} E={E}")
  

  Solution: The ages are: Andrew = 2; Brian = 8; Colin = 3; David = 7; E = 5.

  We don’t know the name of the fifth child, but it may well start with an E.

  Summing the five ages and reversing gives:

  2 + 8 + 3 + 7 + 5 = 25 → 52

  And 52 is divisible by 2, but not 8, 3, 7, 5.

  With the 4 ages B – E:

  8 + 3 + 7 + 5 = 23 → 32

  And 32 is divisible by 8, but not 3, 7, 5.

  With the 3 ages C – E:

  3 + 7 + 5 = 15 → 51

  And 51 is divisible by 3, but not 7, 5.

  With the 2 ages D – E:

  7 + 5 = 12 → 21

  And 21 is divisible by 7, but not 5.

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  Frits 3:31 pm on 2 September 2025 Permalink | Reply

  from enigma import SubstitutedExpression
  
  def check(*s):
    if (rsum := int(str(sum(s))[::-1])) % s[-1] or not (9 < rsum < 100): 
      return False
    return all(rsum % x for x in s[:-1])
    
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
      "check(E, D)",
      "check(E, D, C)",
      "check(E, D, C, B)",
      "check(E, D, C, B, A)",
    ],
    answer="(A, B, C, D, E)",
    env=dict(check=check),
    d2i=dict([(1, "BCDE")]),
    digits=range(1, 10),
    verbose=0,    # use 256 to see the generated code
  )
  
  # print answers
  for ans in p.answers():
    print(f"{', '.join((x + ': ' + y 
            for x, y in zip('ABCDE', [str(a) for a in ans])))}")
  

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  GeoffR 8:37 pm on 2 September 2025 Permalink | Reply

  from itertools import permutations
  
  for a, b, c, d, e in permutations(range(1, 10), 5):
    fg = a + b + c + d + e
    f, g = fg // 10, fg % 10
    if not f > 0 and g > 0:
      continue
    gf = 10*g + f
    if gf % a == 0 and all (gf % x != 0 for x in (b, c, d, e)):
          
      # Ekiminate Andrew
      hi = b + c + d + e
      h, i = hi // 10, hi % 10
      if not h > 0 and i > 0:
        continue
      ih = 10*i + h
          
      # Eliminate Brian
      if ih % b == 0 and all( ih % x != 0 for x in (c, d, e)):
        jk = c + d + e
        j, k = jk // 10, jk % 10
        if not j > 0 and k > 0:
          continue
        kj = 10*k + j
          
        # Eliminate Colin
        if kj % c == 0 and all(kj % x != 0 for x in (d, e)):
          Lm = d + e
          L, m = Lm // 10, Lm % 10
          if not L > 0 and m > 0:
            continue
          mL = 10*m + L
  
          # Eliminate David
          if mL % d == 0 and mL % e != 0: 
            print(f" A = {a}, B = {b}, C = {c}, D = {d}, E = {e}")
  
  # A = 2, B = 8, C = 3, D = 7, E = 5 
  
  

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  Ruud 7:57 am on 4 September 2025 Permalink | Reply

  import istr
  
  def check(ages, i):
      return [sum(ages[i:])[::-1].is_divisible_by(age) for age in ages[i:]] == [True] + (4 - i) * [False]
  
  for ages in istr.permutations("123456789", 5):
      if all(check(ages, i) for i in range(4)):
          print(ages)
  

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  Ruud 2:56 pm on 4 September 2025 Permalink | Reply

  As a one-liner:

  import istr
  
  print(
      [
          ages
          for ages in istr.permutations("123456789", 5)
          if all(all(sum(ages[i:])[::-1].is_divisible_by(ages[j]) == (j == i) for j in range(i, 5)) for i in range(4))
      ]
  )
  

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