Sunday Times Teaser 2675 – Should Old Acquaintance …

by Victor Bryant

I have assigned integer values between 1 and 26 (in some order) to the letters of the alphabet, two specific values being A = 5 and B = 24. The values of words are then determined by the sum of their letter values.

As a result the value of OLD is the average of the values of SHOULD and ACQUAINTANCE. The values of NEW and YEAR are the same and are equal to the value of OLD when its two digits are reversed.

The above information is sufficient to deduce the values of N, E and W.

What is the value of the word NEW?

2 Comments Leave one →

brian gladman permalink

This is a bit slow and would probably be faster if I had used a recursive approach. But it gets the job done.


from itertools import combinations, permutations

# the given assignments
A, B = 5, 24
# the initial assigned values
ltrs_0 = set((A, B))

# We can show that 71 <= OLD <= 74 so the minimum for any 
# of O, L and D is 71 - 25 - 26 = 20. We don't care which 
# value is assigned to which letter so use combinations
for c1 in combinations((20, 21, 22, 23, 25, 26), 3):
  OLD = sum(c1)
  if not 71 <= OLD <= 74:
    continue
  # the digit reversed value of OLD
  rev = 10 * (OLD % 10) + OLD // 10
  O, L, D = c1
  ltrs_1 = ltrs_0 | set(c1)
  
  # we can show that the remaining letter values are less than
  # 14 and E and N have to be chosen from such values
  for p1 in permutations(set(range(1, 14)) - ltrs_1, 2):
    N, E = p1
    ltrs_2 = ltrs_1 | set(p1)
    
    # the order of the values for C and U doesn't matter
    for c2 in combinations(set(range(1, 14)) - ltrs_2, 2):
      C, U = c2
      ltrs_3 = ltrs_2 | set(c2)
      # we can show that the maximum of C + N + U is 18 
      if C + N + U > 18:
        continue

      # the order of the values for H, I, Q, S and T doesn't matter
      for c3 in combinations(set(range(1, 14)) - ltrs_3, 5):
        H, I, Q, S, T = c3
        # check that SHU + ACQUAINTANCE = OLD
        if 3 * A + 2 * sum(p1 + c2) - E + sum(c3) != OLD:
          continue
        ltrs_4 = ltrs_3 | set(c3)

        # find unassigned values for setting W, R and Y
        rest = set(range(1, 27)) - ltrs_4
        W = rev - N - E        
        if W in rest:
          # now assign R and Y (the order doesn't matter)
          for R, Y in combinations(rest - set((W,)), 2):
            # and check that YEAR ie equal to rev (= NEW)
            if  Y + E + A + R == rev:
              print(rev, (A, B, C, D, E, H, I, L, N, O, Q, R, S, T, U, W, Y))

from itertools import combinations, permutations

# the given assignments

A, B = 5, 24

# the initial assigned values

ltrs_0 = set((A, B))

# We can show that 71 <= OLD <= 74 so the minimum for any

# of O, L and D is 71 - 25 - 26 = 20. We don't care which

# value is assigned to which letter so use combinations

for c1 in combinations((20, 21, 22, 23, 25, 26), 3):

OLD = sum(c1)

if not 71 <= OLD <= 74:

continue

# the digit reversed value of OLD

rev = 10 * (OLD % 10) + OLD // 10

O, L, D = c1

ltrs_1 = ltrs_0 | set(c1)

# we can show that the remaining letter values are less than

# 14 and E and N have to be chosen from such values

for p1 in permutations(set(range(1, 14)) - ltrs_1, 2):

N, E = p1

ltrs_2 = ltrs_1 | set(p1)

# the order of the values for C and U doesn't matter

for c2 in combinations(set(range(1, 14)) - ltrs_2, 2):

C, U = c2

ltrs_3 = ltrs_2 | set(c2)

# we can show that the maximum of C + N + U is 18

if C + N + U > 18:

continue

# the order of the values for H, I, Q, S and T doesn't matter

for c3 in combinations(set(range(1, 14)) - ltrs_3, 5):

H, I, Q, S, T = c3

# check that SHU + ACQUAINTANCE = OLD

if 3 * A + 2 * sum(p1 + c2) - E + sum(c3) != OLD:

continue

ltrs_4 = ltrs_3 | set(c3)

# find unassigned values for setting W, R and Y

rest = set(range(1, 27)) - ltrs_4

W = rev - N - E

if W in rest:

# now assign R and Y (the order doesn't matter)

for R, Y in combinations(rest - set((W,)), 2):

# and check that YEAR ie equal to rev (= NEW)

if Y + E + A + R == rev:

print(rev, (A, B, C, D, E, H, I, L, N, O, Q, R, S, T, U, W, Y))

ahmet cetinbudaklar permalink

This is my approach without checking its uniquness:

NEW = YEAR and OLD = 3A + 2(U + N + C) + Q + I + T + E + S + H gives us 71 < OLD < 75 and 17 < NEW < 57 Thus by trial and error we get: N = 3, E = 11, W = 23, Y = 12, A = 5, R = 9, O = 22, L = 25, D = 26, U = 1, C = 2, Q = 4, I = 6, T = 7, S = 8, H = 10. Happy New Year with full of health and happiness!

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Sunday Times Teaser 2675 – Should Old Acquaintance …

by Victor Bryant

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