Sunday Times Teaser 2952 – Silly Slip

by Victor Bryant

Published April 21 2019 (link)

Please help me find my silly slip. I correctly added a five-figure number to a four-figure number to give a six-figure total. Then I tried to substitute letters for digits systematically and I ended up with

SILLY + SLIP = PLEASE

However, in these letters I have made one silly slip, so please find it and then work out what the correct sum was.

What was the correct six-figure numerical answer?

2 Comments Leave one →

Brian Gladman permalink

Using my alphametic sum solver (sadly very slow):


from alphasum import AlphaSum

words = 'SILLY SLIP PLEASE'

l_set = set(words).difference([' '])
if len(l_set) < 10:
  l_set.add('X')

for i, c in enumerate(words):
  if c != ' ':
    for x in l_set:
      a, b, c = (words[:i] + x + words[i + 1:]).split()
      AlphaSum([a, b], c).go()

from alphasum import AlphaSum

words = 'SILLY SLIP PLEASE'

l_set = set(words).difference([' '])

if len(l_set) < 10:

l_set.add('X')

for i, c in enumerate(words):

if c != ' ':

for x in l_set:

a, b, c = (words[:i] + x + words[i + 1:]).split()

AlphaSum([a, b], c).go()

Erling Torkildsen permalink


from itertools import permutations

digits = list(permutations(range(10),5))
c = ["First digit is P", "Second digit is L", "Fifth digit is S", 
     "Third and last digits are equal", "Third digit not in {S,I,L,Y,P}", 
     "Fourth digit not in {S,I,L,Y,P}"]

for S, I, L, Y, P in digits:
  a1 = 10000 * S + 1000 * I + 100 * L + 10 * L + Y
  a2 = 1000 * S + 100 * L + 10 * I + P
  n = str(a1 + a2)

  if len(n) == 6:
    # Tests
    t1 = n[0] == str(P) # First digit is P
    t2 = n[1] == str(L) # Second digit is L
    t3 = n[4] == str(S) # Fifth digit is S
    t4 = n[2] == n[5]   # Third and last digits are equal
    t5 = int(n[2]) not in {S, I, L, Y, P}
    t6 = int(n[3]) not in {S, I, L, Y, P}

    tests = (t1, t2, t3, t4, t5, t6)
    f = 0
    for i, t in enumerate(tests):
      if not t:
        m = c[i]
        f = f + 1
    if f == 1:
      print(f"SILLY + SLIP = PLEASE\n{a1} + {a2} = {n} <-- Violates '{m}'")

from itertools import permutations

digits = list(permutations(range(10),5))

c = ["First digit is P", "Second digit is L", "Fifth digit is S",

"Third and last digits are equal", "Third digit not in {S,I,L,Y,P}",

"Fourth digit not in {S,I,L,Y,P}"]

for S, I, L, Y, P in digits:

a1 = 10000 * S + 1000 * I + 100 * L + 10 * L + Y

a2 = 1000 * S + 100 * L + 10 * I + P

n = str(a1 + a2)

if len(n) == 6:

# Tests

t1 = n[0] == str(P) # First digit is P

t2 = n[1] == str(L) # Second digit is L

t3 = n[4] == str(S) # Fifth digit is S

t4 = n[2] == n[5] # Third and last digits are equal

t5 = int(n[2]) not in {S, I, L, Y, P}

t6 = int(n[3]) not in {S, I, L, Y, P}

tests = (t1, t2, t3, t4, t5, t6)

f = 0

for i, t in enumerate(tests):

if not t:

m = c[i]

f = f + 1

if f == 1:

print(f"SILLY + SLIP = PLEASE\n{a1} + {a2} = {n} <-- Violates '{m}'")

Sunday Times Teaser 2952 – Silly Slip

by Victor Bryant

Published April 21 2019 (link)

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