Sunday Times Teaser 2936 – Multicoloured

by Danny Roth

Published December 30 2018 (link)

George and Martha are selling wall-paper of various colours. By replacing letters with positive digits, they have devised the following multiplication problem:

RED x GREY == YELLOW

The N in GREEN is the remaining positive digit. The red wall-paper is sold only in squares, which is appropriate since RED is a perfect square.

What is the value of GREEN?

3 Comments Leave one →

Brian Gladman permalink


from itertools import permutations

# the nine non-zero digits
digits = set(range(1, 10))

# find the possible values for RED, a three digit
# square with three different non-zero digits
for RED in (x ** 2 for x in range(10, 32)):
  R, E, D = t = [int(x) for x in str(RED)]
  if len(set(t)) != 3:
    continue
  
  # find the possible values of the digits to complete GREY
  for G, Y in permutations(digits.difference(t), 2):
    GREY = 1000 * G + 100 * R + 10 * E + Y

    # check for the correct form of YELLOW
    YELLOW = RED * GREY
    try:
      y, e, L, l, O, W = (int(x) for x in str(YELLOW))
      s8 = {D, E, G, L, O, R, W, Y}
      if (e, l, y) != (E, L, Y) or 0 in s8 or len(s8) != 8:
        continue
    except ValueError:
      continue
    
    # find the remaining non-zero digit for N to complete GREEN 
    N, = sorted(digits.difference(s8))
    GREEN = 10000 * G + 1000 * R + 110 * E + N 
    print(f'GREEN = {GREEN}, RED({RED}) x GREY({GREY}) == YELLOW({YELLOW}).')

from itertools import permutations

# the nine non-zero digits

digits = set(range(1, 10))

# find the possible values for RED, a three digit

# square with three different non-zero digits

for RED in (x ** 2 for x in range(10, 32)):

R, E, D = t = [int(x) for x in str(RED)]

if len(set(t)) != 3:

continue

# find the possible values of the digits to complete GREY

for G, Y in permutations(digits.difference(t), 2):

GREY = 1000 * G + 100 * R + 10 * E + Y

# check for the correct form of YELLOW

YELLOW = RED * GREY

try:

y, e, L, l, O, W = (int(x) for x in str(YELLOW))

s8 = {D, E, G, L, O, R, W, Y}

if (e, l, y) != (E, L, Y) or 0 in s8 or len(s8) != 8:

continue

except ValueError:

continue

# find the remaining non-zero digit for N to complete GREEN

N, = sorted(digits.difference(s8))

GREEN = 10000 * G + 1000 * R + 110 * E + N

print(f'GREEN = {GREEN}, RED({RED}) x GREY({GREY}) == YELLOW({YELLOW}).')

geoffrounce permalink


from itertools import permutations

def is_sq(n): 
  for a in range(2, int(n ** 0.5) + 1): 
    if a * a == n: 
      return True 
  return False

digits = set([1, 2, 3, 4, 5, 6, 7, 8, 9])

# 1st stage permutation
for p1 in permutations (digits, 3):
  r, e, d = p1
  red = 100 * r + 10 * e + d
  if not is_sq(red) : continue

  # 2nd stage permutation
  qs = digits.difference(p1)

  for p2 in permutations(qs, 5):
    y, l, o, w, g = p2
    grey = 1000 * g + 100 * r + 10 * e + y
    yellow = 100000 * y + 10000 * e + 1000 * l + 100 * l + 10 * o + w
    
    if red * grey == yellow: 
      n, = digits.difference(set((r, e, d, y, l, o, w, g)))
      green = 10000 * g + 1000 * r + 100 * e + 10 * e + n
      print( 'Multiplication problem is {} * {} = {}'. format(red, grey, yellow))
      print('GREEN = ', green)

from itertools import permutations

def is_sq(n):

for a in range(2, int(n ** 0.5) + 1):

if a * a == n:

return True

return False

digits = set([1, 2, 3, 4, 5, 6, 7, 8, 9])

# 1st stage permutation

for p1 in permutations (digits, 3):

r, e, d = p1

red = 100 * r + 10 * e + d

if not is_sq(red) : continue

# 2nd stage permutation

qs = digits.difference(p1)

for p2 in permutations(qs, 5):

y, l, o, w, g = p2

grey = 1000 * g + 100 * r + 10 * e + y

yellow = 100000 * y + 10000 * e + 1000 * l + 100 * l + 10 * o + w

if red * grey == yellow:

n, = digits.difference(set((r, e, d, y, l, o, w, g)))

green = 10000 * g + 1000 * r + 100 * e + 10 * e + n

print( 'Multiplication problem is {} * {} = {}'. format(red, grey, yellow))

print('GREEN = ', green)

Jim Randell permalink

This puzzle can easily be solved with the [[ SubstitutedExpression() ]] solver in the enigma.py Python library.

[ https://repl.it/@jim_r/teaser2936 ]

Python

SubstitutedExpression --digits="1-9" --answer="GREEN" "RED * GREY = YELLOW" "is_square(RED)" # optional

1
2
3
4
5
6
7
8
9

SubstitutedExpression

--digits="1-9"
--answer="GREEN"

"RED * GREY = YELLOW"
"is_square(RED)" # optional

The final expression is optional, but provides a small decrease in run time.

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Sunday Times Teaser 2936 – Multicoloured

by Danny Roth

Published December 30 2018 (link)

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