Sunday Times Teaser 2962 – Book Receipts

by Howard Williams

Published June 30 2019 (link)

My wife recently purchased two books from her local bookshop. She showed me the receipt, which showed the cost of each book and the three-figure total cost. I noticed that all of the digits from 1 to 9 had been printed. Coincidentally, exactly the same happened to me when buying two different books, but my more expensive book cost more than hers. In fact, it would not have been possible for that book to have cost more.

How much did I pay for the more expensive book?

6 Comments Leave one →

Brian Gladman permalink

Using my AlphaSum alphametic solver:


from alphasum import AlphaSum

max_v = (0, 0, 0)
# find solutions for the sum 
ss = AlphaSum(['ABC', 'DEF'], 'GHI', digits=range(1, 10))
for s in ss.solve():
  # extract the three values and look for the most expensive book
  a, b, c = sorted(int(ss.substitute(x)) for x in ('ABC', 'DEF', 'GHI'))
  if a < b and b > max_v[1]:
    max_v = (a, b, c)
a, b, c = (x / 100 for x in max_v)
print(f"Howard's most expensive book cost £{b} (£{a} + £{b} = £{c}).")

from alphasum import AlphaSum

max_v = (0, 0, 0)

# find solutions for the sum

ss = AlphaSum(['ABC', 'DEF'], 'GHI', digits=range(1, 10))

for s in ss.solve():

# extract the three values and look for the most expensive book

a, b, c = sorted(int(ss.substitute(x)) for x in ('ABC', 'DEF', 'GHI'))

if a < b and b > max_v[1]:

max_v = (a, b, c)

a, b, c = (x / 100 for x in max_v)

print(f"Howard's most expensive book cost £{b} (£{a} + £{b} = £{c}).")

Here is an alternative version that is faster (30ms) than that above (43ms).


from itertools import combinations, permutations

# define the sum as:
#
#    A B C    lesser priced book
#  + D E F    most expensive book
#    -----
#    G H I    total cost

max_v = (0, 0, 0)
# since A < D < G we can use a combination to guarantee this
for c in combinations(range(1, 10), 3):
  A, D, G = c

  # now permute the remaining digits to complete the book costs
  for p in permutations(set(range(1, 10)).difference(c), 4):
    B, C, E, F = p
  
    ABC = 100 * A + 10 * B + C
    DEF = 100 * D + 10 * E + F
    GHI = ABC + DEF
    g, HI = divmod(GHI, 100)
    # form the set of unallocated digits
    s2 = set(range(1, 10)).difference(c + p)
    
    # check that the top digit of the sum is correct
    # and that the unallocated digits match H and I
    if g == G and set(divmod(HI, 10)) == s2:
      if DEF > max_v[1]:
        max_v = (ABC, DEF, GHI)

a, b, c = (x / 100 for x in max_v)
print(f"Howard's most expensive book cost £{b} (£{a} + £{b} = £{c}).")

from itertools import combinations, permutations

# define the sum as:

# A B C lesser priced book

# + D E F most expensive book

# -----

# G H I total cost

max_v = (0, 0, 0)

# since A < D < G we can use a combination to guarantee this

for c in combinations(range(1, 10), 3):

A, D, G = c

# now permute the remaining digits to complete the book costs

for p in permutations(set(range(1, 10)).difference(c), 4):

B, C, E, F = p

ABC = 100 * A + 10 * B + C

DEF = 100 * D + 10 * E + F

GHI = ABC + DEF

g, HI = divmod(GHI, 100)

# form the set of unallocated digits

s2 = set(range(1, 10)).difference(c + p)

# check that the top digit of the sum is correct

# and that the unallocated digits match H and I

if g == G and set(divmod(HI, 10)) == s2:

if DEF > max_v[1]:

max_v = (ABC, DEF, GHI)

a, b, c = (x / 100 for x in max_v)

print(f"Howard's most expensive book cost £{b} (£{a} + £{b} = £{c}).")

Erling Torkildsen permalink

As no specific currency is given, my answer is not adjusted in any way. (In my local bookshop the prices are in NOK and my answer is about the price of a rather expensive book in that currency.)


from itertools import permutations

a = {1, 2, 3, 4, 5, 6, 7, 8, 9}
b = permutations(a, 3)
maks = 0
for x in b:
  c = permutations(a - set(x), 3)
  for y in c:
    d = permutations(a - set(x + y), 3)
    for z in d:
      X = x[0] * 100 + x[1] * 10 + x[2]
      Y = y[0] * 100 + y[1] * 10 + y[2]
      Z = z[0] * 100 + z[1] * 10 + z[2]
      if X + Y == Z:
        if max(X, Y) > maks:
          maks = max(X, Y)
      if X + Z == Y:
        if max(X, Z) > maks:
          maks = max(X, Z)
      if Y + Z == X:
        if max(Y, Z) > maks:
          maks = max(Y, Z)

print(f"The more expensive book costs {maks}.")

from itertools import permutations

a = {1, 2, 3, 4, 5, 6, 7, 8, 9}

b = permutations(a, 3)

maks = 0

for x in b:

c = permutations(a - set(x), 3)

for y in c:

d = permutations(a - set(x + y), 3)

for z in d:

X = x[0] * 100 + x[1] * 10 + x[2]

Y = y[0] * 100 + y[1] * 10 + y[2]

Z = z[0] * 100 + z[1] * 10 + z[2]

if X + Y == Z:

if max(X, Y) > maks:

maks = max(X, Y)

if X + Z == Y:

if max(X, Z) > maks:

maks = max(X, Z)

if Y + Z == X:

if max(Y, Z) > maks:

maks = max(Y, Z)

print(f"The more expensive book costs {maks}.")

SandyR permalink

Sorry, have just started programming and working on a tablet so don’t know how to paste formatted code. Apologies if inelegant.


import itertools

exp_book_list = []
perm = itertools.permutations('123456789')
for a, b, c, d, e, f, g, h, i in perm:
  
  cheap_book = 100 * int(a) + 10 * int(b) + int(c)
  exp_book = 100 * int(d) + 10 * int(e) + int(f)
  total = 100 * int(g) + 10 * int(h) + int(i)
  
  if exp_book > cheap_book and cheap_book + exp_book == total:
    exp_book_list.append(exp_book)

exp_book_list.sort()
print(f'The most expensive book cost £{exp_book_list[-1] / 100}')

import itertools

exp_book_list = []

perm = itertools.permutations('123456789')

for a, b, c, d, e, f, g, h, i in perm:

cheap_book = 100 * int(a) + 10 * int(b) + int(c)

exp_book = 100 * int(d) + 10 * int(e) + int(f)

total = 100 * int(g) + 10 * int(h) + int(i)

if exp_book > cheap_book and cheap_book + exp_book == total:

exp_book_list.append(exp_book)

exp_book_list.sort()

print(f'The most expensive book cost £{exp_book_list[-1] / 100}')

Brian Gladman permalink

Welcome to the site SanyR! The menu item ‘About’ gives information on how you can post formatted code.

Reply
- SandyR permalink
  
  Thanks Brian, I will check that. Your site has been really useful for me as it helps me put my learning into practice and I can get tips from the answers posted.
  
  Reply

Erling Torkildsen permalink

After a piece of advice from Brian an improved version of my first upload turns out like below. As this site obviously works as a workshop for learners of Python (like myself), I let my first upload stand.


from itertools import permutations

a = set(range(1, 10))

maks = 0

for x in permutations(a, 3):
  X = x[0] * 100 + x[1] * 10 + x[2]
  
  for y in permutations(a - set(x), 3):
    Y = y[0] * 100 + y[1] * 10 + y[2]
    if X > Y:
      continue

    for z in permutations(a - set(x + y), 3):
      Z = z[0] * 100 + z[1] * 10 + z[2]
      if Y > Z:
        continue

      if X + Y == Z and Y > maks:
        maks = Y

print(f"The more expensive book costs {maks}.")

from itertools import permutations

a = set(range(1, 10))

maks = 0

for x in permutations(a, 3):

X = x[0] * 100 + x[1] * 10 + x[2]

for y in permutations(a - set(x), 3):

Y = y[0] * 100 + y[1] * 10 + y[2]

if X > Y:

continue

for z in permutations(a - set(x + y), 3):

Z = z[0] * 100 + z[1] * 10 + z[2]

if Y > Z:

continue

if X + Y == Z and Y > maks:

maks = Y

print(f"The more expensive book costs {maks}.")

Sunday Times Teaser 2962 – Book Receipts

by Howard Williams

Published June 30 2019 (link)

Leave a comment to Brian Gladman Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta