Sunday Times Teasers 3185 – Multiple Squares

by BRG on October 6, 2023

by Howard Williams

Published Sunday October 08 2023 (link)

My grandson likes to compile 3×3 magic squares, where each of the three rows of numbers, each of the three columns of numbers and both of the straight diagonals, add up to the same total. This he did without repeating any of the nine numbers in the square.

He has now progressed to compiling similar 3×3 squares, which instead of the eight rows, columns and diagonals of numbers adding to the same total, they instead multiply to produce the same product. In his first such square this product was 32,768. He was able to find every square of nine different whole numbers that gives this product, excluding identical rotational and mirrored squares.

What, in ascending order were the totals of the nine numbers in each of his different squares?

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BRG permalink


# Each row/colum/diagonal has a product of 32768 (= 2^15) so each value in
# the 3 x 3 square is a power of 2 and these powers make an additive magic 
# square with a common sum of 15. With x < y, y <> 2.x, x + y <= e and the 
# the common sum of 3.e, a 3 x 3 additive magic square is given by:
#
#      e - x   e + x + y   e - y
#    e + x - y     e     e - x + y
#      e + y   e - x - y   e + x

e, sum_pwrs = 5, {}
for y in range(e + 1):
  for x in range(min(y, e - y + 1)):
    if 2 * x != y and len(set((msqr := tuple(2 ** p for p in
                 (e - x, e + x + y, e - y,
                  e + x - y, e, e - x + y,
                  e + y, e - x - y, e + x))))) == 9:
      sum_pwrs[sum(msqr)] = msqr

for k in sorted(sum_pwrs):
  a, b, c, d, e, f, g, h, i = sum_pwrs[k]  
  print(f'[{a:3}{b:6}{c:3} |{d:3}{e:5}{f:5} |{g:4}{h:3}{i:4} ] -> {k}')

# Each row/colum/diagonal has a product of 32768 (= 2^15) so each value in

# the 3 x 3 square is a power of 2 and these powers make an additive magic

# square with a common sum of 15. With x < y, y <> 2.x, x + y <= e and the

# the common sum of 3.e, a 3 x 3 additive magic square is given by:

# e - x e + x + y e - y

# e + x - y e e - x + y

# e + y e - x - y e + x

e, sum_pwrs = 5, {}

for y in range(e + 1):

for x in range(min(y, e - y + 1)):

if 2 * x != y and len(set((msqr := tuple(2 ** p for p in

(e - x, e + x + y, e - y,

e + x - y, e, e - x + y,

e + y, e - x - y, e + x))))) == 9:

sum_pwrs[sum(msqr)] = msqr

for k in sorted(sum_pwrs):

a, b, c, d, e, f, g, h, i = sum_pwrs[k]

print(f'[{a:3}{b:6}{c:3} |{d:3}{e:5}{f:5} |{g:4}{h:3}{i:4} ] -> {k}')

An alternative formulation:


from number_theory import divisor_tuples

# A multiplicative 3 x 3 magic square can be expressed in the form:
# 
#     a.c    a.b^2.c^2   a.b
#    a.b^2     a.b.c    a.c^2
#   a.b.c^2      a     a.b^2.c
#
# with the common product (a.b.c)^3, which in our case equals 32768

print(sorted((sum(ms) for a, b, c in divisor_tuples(32, 3) if len(set((
        ms := (a * c, a * b * b * c * c, a * b, a * b * b, a * b * c,
               a * c * c, a * b * c * c, a, a * b * b * c)))) == 9)))

from number_theory import divisor_tuples

# A multiplicative 3 x 3 magic square can be expressed in the form:

# a.c a.b^2.c^2 a.b

# a.b^2 a.b.c a.c^2

# a.b.c^2 a a.b^2.c

# with the common product (a.b.c)^3, which in our case equals 32768

print(sorted((sum(ms) for a, b, c in divisor_tuples(32, 3) if len(set((

ms := (a * c, a * b * b * c * c, a * b, a * b * b, a * b * c,

a * c * c, a * b * c * c, a, a * b * b * c)))) == 9)))

Tony Smith permalink

Brian
It hardly matters but there is a typo in R2C3 of your yellow additive square.

Reply
- BRG permalink
  
  Thanks Tony, corrected now (with other corrections).
  
  Reply

Frits permalink

One liner for Brian’s first program. Always 2 numbers between (and including) 0 and 10 must be missing in the magic square.

Developed in Thailand on an Android smartphone with the app “Coding Python”.


print(sorted(sum(2**i for i in range(11)) \
      - 2**(m := set(range(5)).difference([e - x, e - y, e - x - y, e - y + x]).pop()) \
      - 2**(10 - m) \
  for e in [5] \
  for x in range(1, e)
  for y in [y for y in range(x + 1, e - x + 1) if y != 2 * x]))

print(sorted(sum(2**i for i in range(11)) \

- 2**(m := set(range(5)).difference([e - x, e - y, e - x - y, e - y + x]).pop()) \

- 2**(10 - m) \

for e in [5] \

for x in range(1, e)

for y in [y for y in range(x + 1, e - x + 1) if y != 2 * x]))

Sunday Times Teasers 3185 – Multiple Squares

by Howard Williams

Published Sunday October 08 2023 (link)

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