Sunday Times Teaser 2636 – Simple Easter Teaser

by Andrew Skidmore

Different letters have been consistently used to represent the different digits in three numbers giving the words:

SIMPLE EASTER TEASER

One of the numbers is the sum of the other two and one or more of them is odd.

What number does EASTER represent?

3 Comments Leave one →

brian gladman permalink

A bit messy but it gets the job done.


from itertools import permutations as perm

# permute the digits for the letters in EASTER and TEASER 
for E, A, S, T, R in perm(range(10), 5):
  # E, S and T are non-zero and either E, R or both must be odd
  if 0 not in (S, E, T) and (E % 2 or R % 2):
    # form EASTER and TEASER in their numeric form 
    eas, er = 100 * E + 10 * A + S, 10 * E + R
    easter = 1000 * eas + 100 * T + er
    teaser = 100000 * T + 100 * eas + er
    # now permute the remaining digits for the other letters
    for I, M, P, L in perm(set(range(10)) - set((E, A, S, T, R)), 4):
      # and form SIMPLE in numeric form
      simple = 100000 * S + 10000 * I + 1000 * M + 100 * P + 10 * L + E
      t = (simple, easter, teaser)
      # check that one number is the sum of the other two
      if sum(t) == 2 * max(t):
        # output the result
          fs = 'EASTER: {0[1]:d} (SIMPLE: {0[0]:d}, TEASER: {0[2]:d}).'
          print(fs.format(t))

from itertools import permutations as perm

# permute the digits for the letters in EASTER and TEASER

for E, A, S, T, R in perm(range(10), 5):

# E, S and T are non-zero and either E, R or both must be odd

if 0 not in (S, E, T) and (E % 2 or R % 2):

# form EASTER and TEASER in their numeric form

eas, er = 100 * E + 10 * A + S, 10 * E + R

easter = 1000 * eas + 100 * T + er

teaser = 100000 * T + 100 * eas + er

# now permute the remaining digits for the other letters

for I, M, P, L in perm(set(range(10)) - set((E, A, S, T, R)), 4):

# and form SIMPLE in numeric form

simple = 100000 * S + 10000 * I + 1000 * M + 100 * P + 10 * L + E

t = (simple, easter, teaser)

# check that one number is the sum of the other two

if sum(t) == 2 * max(t):

# output the result

fs = 'EASTER: {0[1]:d} (SIMPLE: {0[0]:d}, TEASER: {0[2]:d}).'

print(fs.format(t))

geoffrounce permalink

Here is my solution:


from itertools import permutations

for p in permutations((1, 2, 3, 4, 5, 6, 7, 8, 9, 0), 9):
  s, i, m, p, l, e, a, t, r = p
  if all(y != 0 for y in (s, e, t)):
    simple = e + 10*l + 100*p + 1000*m + 10000*i + 100000*s
    easter = r + 10*e + 100*t + 1000*s + 10000*a + 100000*e
    teaser = r + 10*e + 100*s + 1000*a + 10000*e + 100000*t
    if any(x % 2 == 1 for x in (simple, easter, teaser)):
      if (simple == easter + teaser or 
          easter == simple + teaser or
          teaser == simple + easter):
        print('simple easter teaser = ',simple, easter, teaser)

from itertools import permutations

for p in permutations((1, 2, 3, 4, 5, 6, 7, 8, 9, 0), 9):

s, i, m, p, l, e, a, t, r = p

if all(y != 0 for y in (s, e, t)):

simple = e + 10*l + 100*p + 1000*m + 10000*i + 100000*s

easter = r + 10*e + 100*t + 1000*s + 10000*a + 100000*e

teaser = r + 10*e + 100*s + 1000*a + 10000*e + 100000*t

if any(x % 2 == 1 for x in (simple, easter, teaser)):

if (simple == easter + teaser or

easter == simple + teaser or

teaser == simple + easter):

print('simple easter teaser = ',simple, easter, teaser)

ahmet cetinbudaklar permalink

My analytical approach involved to start with:
itSIMPLE=EASTER+TEASER , taking R’s as odd digits and E as an even digit which helped me to reach the answer quite easily

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Sunday Times Teaser 2636 – Simple Easter Teaser

by Andrew Skidmore

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