Sunday Times Teaser 2726 – Christmas Star

by David Buontempo

I wanted to place the digits 1 to 10 on the points of a Christmas star where its five lines cross or join in such a way that the sums of the four digits on each of its lines were the same. I couldn’t do this so I used the digits 1 to 9 instead with one digit used twice. One line in the result did not include the repeated digit.

What were the four numbers on this line?

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brian gladman permalink


from itertools import permutations

# Organise the variables in the Christmas Star thus:
# 
#           A
#     E   F   G   B
#       J       H
#           I
#      D         C
#
# These sums are hence all equal (to, say, S): 
#
#    A + C + G + H  =  B + D + H + I  =  C + E + I + J  = 
#    D + A + J + F  =  E + B + F + G  =  S
#
# Adding these and using N for the duplicated digit we have:
# 
#    5.S = 2.(A + ... + J) where A .. J are in (1, ..., 9, N), so
#    5.S = 2.(45 + N)  --> S = 18 + 2.N / 5  --> N = 5 and S = 20 
#
# F, G, H, I and J can now be expressed in terms of A, B, C, D, E:
#
#   F = 10 - B + C - D,  G = 10 - C + D - E,  H = 10 - D + E - A,
#   I = 10 - E + A - B,  J = 10 - A + B - C

# the digits to be used
digits = sorted(tuple(range(1, 10)) + (5,))
# a set for storing solutions
sol = set()
# choose the five digits for the outer ring
for otr in permutations(digits, 5):
  # remove rotated and reflected solutiuons
  if otr[0] == min(otr) and otr[1] < otr[4]:
    # now compute the values in the inner ring
    inr = tuple((10 - otr[(i + 1) % 5] + otr[(i + 2) % 5]
                    - otr[(i + 3) % 5]) for i in range(5))
    # and check that all the digits are used
    if sorted(otr + inr) == digits:
      # compile the values on the five lines
      lines = tuple((otr[i], otr[(i + 2) % 5], inr[(i + 1) % 5], 
                     inr[(i + 2) % 5]) for i in range(5))
      # and find the line on which five is absent
      l = [sorted(x) for x in lines if 5 not in x]
      sol.add((otr, inr, tuple(l[0])))

# output the solutions
for otr, inr, lne in sorted(sol):
  print('outer {}, inner {} -> {}'.format(otr, inr, lne))

from itertools import permutations

# Organise the variables in the Christmas Star thus:

# A

# E F G B

# J H

# I

# D C

# These sums are hence all equal (to, say, S):

# A + C + G + H = B + D + H + I = C + E + I + J =

# D + A + J + F = E + B + F + G = S

# Adding these and using N for the duplicated digit we have:

# 5.S = 2.(A + ... + J) where A .. J are in (1, ..., 9, N), so

# 5.S = 2.(45 + N) --> S = 18 + 2.N / 5 --> N = 5 and S = 20

# F, G, H, I and J can now be expressed in terms of A, B, C, D, E:

# F = 10 - B + C - D, G = 10 - C + D - E, H = 10 - D + E - A,

# I = 10 - E + A - B, J = 10 - A + B - C

# the digits to be used

digits = sorted(tuple(range(1, 10)) + (5,))

# a set for storing solutions

sol = set()

# choose the five digits for the outer ring

for otr in permutations(digits, 5):

# remove rotated and reflected solutiuons

if otr[0] == min(otr) and otr[1] < otr[4]:

# now compute the values in the inner ring

inr = tuple((10 - otr[(i + 1) % 5] + otr[(i + 2) % 5]

- otr[(i + 3) % 5]) for i in range(5))

# and check that all the digits are used

if sorted(otr + inr) == digits:

# compile the values on the five lines

lines = tuple((otr[i], otr[(i + 2) % 5], inr[(i + 1) % 5],

inr[(i + 2) % 5]) for i in range(5))

# and find the line on which five is absent

l = [sorted(x) for x in lines if 5 not in x]

sol.add((otr, inr, tuple(l[0])))

# output the solutions

for otr, inr, lne in sorted(sol):

print('outer {}, inner {} -> {}'.format(otr, inr, lne))

Sunday Times Teaser 2726 – Christmas Star

by David Buontempo

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