Sunday Times Teaser 2783 – Old Boys’ Banquet

by Danny Roth

Published: 24 January 2016 (link)

George and Martha have arranged the seating plan for the annual Old Boys banquet; it involves a number of tables, each seating 50 diners. More than half the Old Boys are bringing one female guest each and the rest are coming alone. Martha wrote down three numbers, namely the number of Old Boys bringing a guest, the number of Old Boys coming alone, and the total number of Old Boys coming. George noted that the three numbers between them used each of the digits 0 to 9 exactly once.

How many Old Boys are bringing a guest, and how many are coming alone?

5 Comments Leave one →

Brian Gladman permalink


# the number of boys with guests
for bg in range(100, 10000):
  sbg = str(bg)
  # ensure that its digits are different
  if len(sbg) == len(set(sbg)):
    # try numbers of boys alone that makes the 
    # boy plus girl total a multiple of fifty 
    for b in range((-2 * bg) % 50, bg, 50):
      # ensure that the three numbers have exactly
      # nine different digits in total
      s = sbg + str(b) + str(bg + b) 
      if len(s) == len(set(s)) == 10:
        fs = 'Boys with a guest {}, boys alone {} (boys {}, all {}).'
        print(fs.format(bg, b, bg + b, 2 * bg + b))

# the number of boys with guests

for bg in range(100, 10000):

sbg = str(bg)

# ensure that its digits are different

if len(sbg) == len(set(sbg)):

# try numbers of boys alone that makes the

# boy plus girl total a multiple of fifty

for b in range((-2 * bg) % 50, bg, 50):

# ensure that the three numbers have exactly

# nine different digits in total

s = sbg + str(b) + str(bg + b)

if len(s) == len(set(s)) == 10:

fs = 'Boys with a guest {}, boys alone {} (boys {}, all {}).'

print(fs.format(bg, b, bg + b, 2 * bg + b))

A bit of analysis provides a much faster solution:


# p = pairs, b = boys alone, j and k integer
# 
#   2p + b = 50j  ;   p + b =  9k
#
#   b = 18k - 50j ;   k >= 25j / 9
#   p = 50j -  9k ;   k <= 50j / 9

for j in range(1, 240):
  for k in range(25 * j // 9, 50 * (j + 8) // 9 + 1):
    p = 50 * j - 9 * k
    b = 18 * k - 50 * j
    if p > b > 0:
      s = str(p) + str(b) + str(p + b)
      if len(s) == len(set(s)) == 10:
        fs = 'Boys with a guest {}, boys alone {} (boys {}, all {}).'
        print(fs.format(p, b, p + b, 2 * p + b))

# p = pairs, b = boys alone, j and k integer

# 2p + b = 50j ; p + b = 9k

# b = 18k - 50j ; k >= 25j / 9

# p = 50j - 9k ; k <= 50j / 9

for j in range(1, 240):

for k in range(25 * j // 9, 50 * (j + 8) // 9 + 1):

p = 50 * j - 9 * k

b = 18 * k - 50 * j

if p > b > 0:

s = str(p) + str(b) + str(p + b)

if len(s) == len(set(s)) == 10:

fs = 'Boys with a guest {}, boys alone {} (boys {}, all {}).'

print(fs.format(p, b, p + b, 2 * p + b))

kevin Fox permalink

Using the code below I seems to get two valid answers


from math import floor

def AllDifferent(List):
  Flag = True
  if len(List) != 10:
    Flag = False
  List.sort()
  for x in range(1, len(List)):
    if List[x - 1] == List[x]:
      Flag = False
  return(Flag)

def main():
  for PlusGuest in range(100, 900):
    for OnOwn in range (100, 900):
      Total = 2 * PlusGuest + OnOwn
      if Total / 50 == floor(Total / 50):
        List = []
        temp = str(PlusGuest)
        for x in range(len(temp)):
          List.append(temp[x - 1])

        temp = str(OnOwn)
        for x in range(len(temp)):
          List.append(temp[x - 1])

        temp = str(Total)
        for x in range(len(temp)):
          List.append(temp[x - 1])

        if PlusGuest > OnOwn:
          if (AllDifferent(List)) == True:
            print(PlusGuest, OnOwn, Total)
  print('End')
main()

from math import floor

def AllDifferent(List):

Flag = True

if len(List) != 10:

Flag = False

List.sort()

for x in range(1, len(List)):

if List[x - 1] == List[x]:

Flag = False

return(Flag)

def main():

for PlusGuest in range(100, 900):

for OnOwn in range (100, 900):

Total = 2 * PlusGuest + OnOwn

if Total / 50 == floor(Total / 50):

List = []

temp = str(PlusGuest)

for x in range(len(temp)):

List.append(temp[x - 1])

temp = str(OnOwn)

for x in range(len(temp)):

List.append(temp[x - 1])

temp = str(Total)

for x in range(len(temp)):

List.append(temp[x - 1])

if PlusGuest > OnOwn:

if (AllDifferent(List)) == True:

print(PlusGuest, OnOwn, Total)

print('End')

main()

Brian Gladman permalink

Welcome to the site Kevin! You are right that this teaser has multiple solutions (four in fact) but there is a problem somewhere in your code since neither of your solutions is correct – the three numbers involve duplicate digits: 732 + 486 = 1218 and 793 + 264 = 1057. It seems that your Total includes the girls but the total that is needed in finding the ten digits is the total number of boys.

I am guessing that you are fairly new to Python and, perhaps, more used to coding in C? Once you get to grips with Python you will find that it has many nice features!

By the way, to get your code formatted you need to add tags (see the menu item ‘About’ above).

Reply

Jim Randell permalink

Here’s a solution that uses the SubstitutedSum() solver from the enigma.py


from enigma import SubstitutedSum, printf

# b0 = old boys without guests
# b1 = old boys with a guest
# consider the pandigital sum: b0 + b1 = b
# it can be: AB + CDEF = GHIJ or ABC + DEF = GHIJ

# solve the pandigital sum
digits = 'ABCDEFGHIJ'
for k in (2, 3):
  (x, y, z) = (digits[0:k], digits[k:6], digits[6:])
  p = SubstitutedSum([x, y], z)
  for s in p.solve():
    # turn the solution into numbers
    (b0, b1, b) = (int(p.substitute(s, w)) for w in (x, y, z))
    # more than half the old boys have a +1
    if not(b1 > b0): continue
    # so the total number of people is...
    t = b + b1
    # and the number of tables is...
    (n, r) = divmod(t, 50)
    if r > 0: continue
  
    printf("b0={b0} b1={b1} boys={b}, people={t} tables={n}")

from enigma import SubstitutedSum, printf

# b0 = old boys without guests

# b1 = old boys with a guest

# consider the pandigital sum: b0 + b1 = b

# it can be: AB + CDEF = GHIJ or ABC + DEF = GHIJ

# solve the pandigital sum

digits = 'ABCDEFGHIJ'

for k in (2, 3):

(x, y, z) = (digits[0:k], digits[k:6], digits[6:])

p = SubstitutedSum([x, y], z)

for s in p.solve():

# turn the solution into numbers

(b0, b1, b) = (int(p.substitute(s, w)) for w in (x, y, z))

# more than half the old boys have a +1

if not(b1 > b0): continue

# so the total number of people is...

t = b + b1

# and the number of tables is...

(n, r) = divmod(t, 50)

if r > 0: continue

printf("b0={b0} b1={b1} boys={b}, people={t} tables={n}")

Trevor Kirsten permalink

I took a short cut using Brian’s letter/number substitution Python program on his ‘paper and pencil’ teaser site. If \(efg\), \(hij\) and \(abcd\) are the number of boys with and without partners and the total respectively, then: \(efg + hij = abcd\) where \(efg > hij\). The total number of diners is \(2efg + hij\) which equals \(50N\) where \(N\) is the number of tables and \(2g+j=10\). This gave 2 possible solutions of which only \(752 + 346 = 1098\) satisfied (i). (1950 guests, 39 tables.) Putting 2g + j = 20 gave no solution. Similarly putting \(efgh\) and \(ij\) as the number of boys and \(2h + j\) equal to 10 or 20 quickly gave the other 3 solutions.

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Sunday Times Teaser 2783 – Old Boys’ Banquet

by Danny Roth

Published: 24 January 2016 (link)

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