Enigmatic Code

Programming Enigma Puzzles

Enigma 16: Four, five, six, seven

5 Comments Posted by on 12 December 2012

From New Scientist #1158, 7th June 1979 [link] [link]

“A pattern, that’s what I like”, said Uncle Bungle. “I can’t stand a puzzle that is all over the place. But in this long division sum the four figures that are given have the great merit of being consecutive”. I could see my uncle’s point all right, but unfortunately (and this will not I think surprise my readers) his achievements did not come up to his hopes and expectations. I can hardly bear to tell my readers this, but in fact one of those figures was wrong.

The puzzle, as Uncle Bungle gave it to me, looked like this:

Enigma 16

Which figure was wrong? Find the correct division sum.

[enigma16]

enigma

5 responses to “Enigma 16: Four, five, six, seven”

  1. Jim Randell 12 December 2012 at 12:05 am

    The following Python program runs in 45ms.

    from enigma import irange, printf

# how many of the items in the iterable are False
def check(*i):
  return sum(1 for x in i if not(x))

# we need a * b = c where a is a 2-digit number, b is a 3-digit number
# (with a 0 in the middle), and c is a 4-digit number
for a in irange(10, 99):
  for b3 in irange(0, 9):
    for b1 in irange(1, 9):
      b = b1 * 100 + b3
      c = a * b
      if not(c < 10000): break
      # check the digits filled out
      a1 = a // 10
      c1 = c // 1000
      # at most one of these can be false
      (s1, s2) = ((a1 == 4), (c1 == 6))
      if check(s1, s2) > 1: continue
      # and check the intermediate values
      d = a * b1
      if not(9 < d < 100): continue
      d2 = d % 10
      # at most one of these can be false
      s3 = (d2 == 7)
      if check(s1, s2, s3) > 1: continue
      e1 = (c // 100) - d
      if not(0 < e1 < 10): continue
      # exactly one of the these has to be false
      s4 = (e1 == 5)
      if check(s1, s2, s3, s4) != 1: continue

      printf("{a} x {b} = {c} [4:{s1}, 6:{s2}, 7:{s3}, 5:{s4}]")

    Solution: The figure 4 is wrong, it should be a 5.

    The actual division sum is:

    Enigma 16 - Solution

    Reply
  2. Jim Randell 13 July 2014 at 10:34 am

    Here’s a solution using the [[ SubstitutedDivision() ]] solver from the enigma.py library. It runs in 49ms.

    from enigma import SubstitutedDivision, printf

# the digits in the sum
digits = list("4567")

# replacement digits
replacements = list("4567X")

# make a new array of digits with one of them replaced
for (i, d) in enumerate(digits):
  for x in replacements:
    if x == d: continue
    ds = list(digits)
    ds[i] = x

    # create the sum
    (d4, d5, d6, d7) = ds    
    p = SubstitutedDivision(
      d6 + '???', d4 + '?', '???',
      [('?' + d7, '?'), None, (d5 + '??', '')],
      dict((x, int(x)) for x in set(ds) if x != 'X')
    )

    # solve it
    for s in p.solve():
      printf("{d} -> {x}")
      p.solution(s)
    Reply
  3. Jim Randell 20 April 2017 at 4:25 pm

    Here’s a solution using the [[ SubstitutedExpression() ]] general alphametic solver from the enigma.py library.

    We don’t need to write a program, just provide the expressions and options as arguments, here is the relevant run file:

    #!/usr/bin/env python -m enigma -r

# writing the sum as:
#
#          G H I
#      ---------
#  E F ) A B C D
#        K L
#        ---
#          M C D
#          M C D
#          =====
#
# we see that H = 0, and exactly one of (E != 4, A != 6, L != 7, M != 5)

# use the alphametic solver
SubstitutedExpression

# letters do not necessarily stand for distinct digits
--distinct=""
# assign H=0
--assign="H,0"

# the main division sum
"GHI * EF = ABCD"

# the multiples
"EF * G = KL"
"EF * I = MCD"

# the intermediate subtractions
"AB - KL = M"

# exactly one of these inequalities holds
"sum([E != 4, A != 6, L != 7, M != 5]) = 1"

    It executes in 96ms:

    % python -m enigma -r enigma16.run
(GHI * EF = ABCD) (EF * G = KL) (EF * I = MCD) (AB - KL = M) (sum([E != 4, A != 6, L != 7, M != 5]) = 1)
(109 * 57 = 6213) (57 * 1 = 57) (57 * 9 = 513) (62 - 57 = 5) (sum([5 != 4, 6 != 6, 7 != 7, 5 != 5]) = 1) / A=6 B=2 C=1 D=3 E=5 F=7 G=1 H=0 I=9 K=5 L=7 M=5
[1 solution]
    Reply
  4. Jim Randell 19 July 2017 at 11:28 am

    The new [[ SubstitutedDivision() ]] solver in the enigma.py library is based on the [[ SubstitutedExpression() ]] solver, so we can have an even simpler run file.

    This run file executes in 108ms.

    #!/usr/bin/env python -m enigma -r

# writing the sum as:
#
#          ? ? ?
#      ---------
#  A ? ) C ? ? ?
#        ? D
#        ---
#          ? ? ?
#          B ? ?
#          =====
#
# one of the following is wrong:
#
#  A = 4, B = 5, C = 6, D = 7

SubstitutedDivision

# substituted digits are not necessarily distinct
--distinct=""

# the main division sum
"C??? / A? = ???"

# the intermediate subtraction sums (the second is empty)
"C? - ?D = ?"
""
"??? - B?? = 0"

# exactly one the following inequalities holds
--extra="sum([A != 4, B != 5, C != 6, D != 7]) = 1"
    Reply
  5. geoffrounce 23 November 2017 at 9:34 pm 
    % A Solution in MiniZinc
include "globals.mzn";

var 1..9:A;   var 0..9:B;  var 0..9:C;   % same letters as given solution
var 0..9:D;   var 1..9:E;  var 0..9:F;
var 1..9:G;   var 0..9:H;  var 0..9:I;
var 1..9:K;   var 0..9:L;  var 1..9:M;

var 1000..9999: ABCD = 1000*A +100*B +10*C + D;
var 100..999 : MCD = 100*M + 10*C + D;
var 100..999 : GHI = 100*G + 10*H + I;

var 10..99 : EF = 10 * E + F;
var 10..99 : KL = 10 * K + L;
var 10..99 : AB = 10 * A + B;

constraint GHI * EF = ABCD /\ EF * G = KL /\ EF * I = MCD
/\ AB - KL = M ;

% One digit of 4, 5, 6, or 7 is incorrect
constraint sum([E == 4, A == 6, L == 7, M == 5]) = 3;

solve satisfy;

output [ show(GHI) ++ " x " ++ show(EF) ++ " = " ++ show(ABCD) ] ++ 
 ["\n" ++ show("A, B, C, D, E, F, G, H, I, K, L, M") ++ " = " 
 ++ show([A, B, C, D, E, F, G, H, I, K, L, M]) ];

% 109 x 57 = 6213
%   [ A, B, C, D, E, F, G, H, I, K, L, M ] 
% =  [6, 2, 1, 3, 5, 7, 1, 0, 9, 5, 7, 5]
% ----------
% Finished in 58msec
%
% Reconstructed division sum from above arrays:
%           G H I           1 0 9
%       ---------      ----------
%   E F ) A B C D      57)6 2 1 3
%         K L             5 7
%         ---             ---
%           M C D           5 1 3
%           M C D           5 1 3
%           -----           -----
% Answer : The digit '4' in the divisor is incorrect and should be 5
    Reply

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