New Scientist Enigma 494 – Look Blank

by Susan Denham

From Issue #1646, 7th January 1989

In the following long division sum I’ve marked the position of each digit. I can tell you that there were no 1s and no 0s but that all other digits occurred at least twice.

Also (although you don’t actually need this information) all four digits of the answer were different.

What is the six-figure dividend?

One Comment Leave one →

Brian Gladman permalink

I am using Jim Randell’s Python interface to the MiniZinc constraint solver (available at here) to solve this enigma:

from minizinc import MiniZinc, var, alphametic, substitute

# the MiniZinc model with the array of all symbols in
# the division sum to be completed later
model = f"""

include "globals.mzn";

% declare the decision variables
{var("2..9", "ABCDEFGHIJKLMNOPQRSTUVWX")};
% the array of all symbols in the sum (including repeats) 
{var("array[1..#] of", "2..9", "Y")} = [$];

% overall division sum
constraint {alphametic("{EF} * {ABCD} = {GHIJKL}")};

% partial products
constraint {alphametic("{A} * {EF} = {MNO}")};
constraint {alphametic("{B} * {EF} = {RST}")};
constraint {alphametic("{C} * {EF} = {VW}")};
constraint {alphametic("{D} * {EF} = {XL}")};

% partial differences
constraint {alphametic("{GHI} - {MNO} = {PQ}")};
constraint {alphametic("{PQJ} - {RST} = {U}")};
constraint {alphametic("{UK} - {VW} = {X}")};

% at least two digits from each of 2..9 in the division sum
constraint forall (x in 2..9) (at_least(2, Y, x));

% alternative to the above
% array[1..8] of var 0..9: NC = global_cardinality(Y, [2, 3, 4, 5, 6, 7, 8, 9]);
% constraint forall (x in NC) (x > 1);

solve satisfy;
"""

# the symbols used in the division sum
fmt = """
              A B C D
        -------------
    E F ) G H I J K L
          M N O 
          -----
            P Q J 
            R S T
            -----
                U K 
                V W
                ---
                  X L 
                  X L
                  ===
"""

# find the alphabetic characters in the division sum
chrs = sorted(c for c in fmt if c.isalpha())

# find the points to insert the MiniZinc array (length 
# and content) used to count the number of each digit
ph, pd = model.find('#'), model.find('$')

# update the model to add the character array to it
model = (model[:ph] + str(len(chrs)) + model[ph + 1: pd]
         + f"{','.join(chrs)}" + model[pd+1:])

p = MiniZinc(model)
for s in p.solve():
  print(p.substitute(s, fmt))

from minizinc import MiniZinc, var, alphametic, substitute

# the MiniZinc model with the array of all symbols in

# the division sum to be completed later

model = f"""

include "globals.mzn";

% declare the decision variables

{var("2..9", "ABCDEFGHIJKLMNOPQRSTUVWX")};

% the array of all symbols in the sum (including repeats)

{var("array[1..#] of", "2..9", "Y")} = [$];

% overall division sum

constraint {alphametic("{EF} * {ABCD} = {GHIJKL}")};

% partial products

constraint {alphametic("{A} * {EF} = {MNO}")};

constraint {alphametic("{B} * {EF} = {RST}")};

constraint {alphametic("{C} * {EF} = {VW}")};

constraint {alphametic("{D} * {EF} = {XL}")};

% partial differences

constraint {alphametic("{GHI} - {MNO} = {PQ}")};

constraint {alphametic("{PQJ} - {RST} = {U}")};

constraint {alphametic("{UK} - {VW} = {X}")};

% at least two digits from each of 2..9 in the division sum

constraint forall (x in 2..9) (at_least(2, Y, x));

% alternative to the above

% array[1..8] of var 0..9: NC = global_cardinality(Y, [2, 3, 4, 5, 6, 7, 8, 9]);

% constraint forall (x in NC) (x > 1);

solve satisfy;

"""

# the symbols used in the division sum

fmt = """

A B C D

-------------

E F ) G H I J K L

M N O

-----

P Q J

R S T

-----

U K

V W

---

X L

===

"""

# find the alphabetic characters in the division sum

chrs = sorted(c for c in fmt if c.isalpha())

# find the points to insert the MiniZinc array (length

# and content) used to count the number of each digit

ph, pd = model.find('#'), model.find('$')

# update the model to add the character array to it

model = (model[:ph] + str(len(chrs)) + model[ph + 1: pd]

+ f"{','.join(chrs)}" + model[pd+1:])

p = MiniZinc(model)

for s in p.solve():

print(p.substitute(s, fmt))

New Scientist Enigma 494 – Look Blank

by Susan Denham

From Issue #1646, 7th January 1989

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