Sunday Times Teaser 3022 – Sporty Set

by Victor Bryant

Published Sunday August 23 2020 (link)

There are 100 members of my sports club where we can play tennis, badminton, squash and table tennis (with table tennis being the least popular). Last week I reported to the secretary the numbers who participate in each of the four sports. The digits used overall in the four numbers were different and not zero.

The secretary wondered how many of the members were keen enough to play all four sports, but of course he was unable to work out that number from the four numbers I had given him. However, he used the four numbers to work out the minimum and the maximum possible numbers playing all four sports. His two answers were two-figure numbers, one being a multiple of the other.

How many played table tennis?

2 Comments Leave one →

Brian Gladman permalink


from itertools import permutations

# If N(X) is the number of players of sport X, considering any two
# sports X and Y, we have:
# 
#    N(X) + N(Y) - N(X, Y) <= 100  ==> N(X, Y) >= N(X) + N(Y) - 100
#
# where N(X, Y) is the number playing both. Since both N(X) and N(Y)
# include N(X, Y), the latter is subtracted to avoid double counting.
#
# Considering four sports A, B, C, D we now have:
#
#   N(A, B) >= N(A) + N(B) - 100
#   N(C, D) >= N(C) + N(D) - 100
#   N(A, B, C, D) >= N(A, B) + N(C, D) - 100
#
# giving the minimum number who play all four sports as:
# 
#   N(A, B, C, D) >= N(A) + N(B) + N(C) + N(D) - 300

# use the digits in <dgts> to form <nbr> two digit numbers
# in increasing order and with all digits different (and 
# the sum of the numbers 300 or more)
def nbrs(dgts, nbr, s=()):
  if not nbr:
    if sum(s) >= 300:    
      yield s
  else:
    for p in permutations(dgts, 2):
      n = 10 * p[0] + p[1]
      if not s or n > s[-1]:
        yield from nbrs(dgts.difference(p), nbr - 1, s + (n,))

# for saving solutions 
sol = set()
# consider possible numbers playing each sport
for s in nbrs(set(range(1, 10)), 4):
  # calculate the minimum and maximum numbers
  # that could play all four sports
  mn, mx = max(0, sum(s) - 300), s[0]
  # both have two digits, one a multiple of the other
  if mn >= 10 and mx % mn == 0:
    sol.add((mn, mx))
    print(s)

# output solutions
for mn, mx in sol:
  print(f"{mx} play Table Tennis, a minimum of {mn} play all four.")

from itertools import permutations

# If N(X) is the number of players of sport X, considering any two

# sports X and Y, we have:

# N(X) + N(Y) - N(X, Y) <= 100 ==> N(X, Y) >= N(X) + N(Y) - 100

# where N(X, Y) is the number playing both. Since both N(X) and N(Y)

# include N(X, Y), the latter is subtracted to avoid double counting.

# Considering four sports A, B, C, D we now have:

# N(A, B) >= N(A) + N(B) - 100

# N(C, D) >= N(C) + N(D) - 100

# N(A, B, C, D) >= N(A, B) + N(C, D) - 100

# giving the minimum number who play all four sports as:

# N(A, B, C, D) >= N(A) + N(B) + N(C) + N(D) - 300

# use the digits in <dgts> to form <nbr> two digit numbers

# in increasing order and with all digits different (and

# the sum of the numbers 300 or more)

def nbrs(dgts, nbr, s=()):

if not nbr:

if sum(s) >= 300:

yield s

else:

for p in permutations(dgts, 2):

n = 10 * p[0] + p[1]

if not s or n > s[-1]:

yield from nbrs(dgts.difference(p), nbr - 1, s + (n,))

# for saving solutions

sol = set()

# consider possible numbers playing each sport

for s in nbrs(set(range(1, 10)), 4):

# calculate the minimum and maximum numbers

# that could play all four sports

mn, mx = max(0, sum(s) - 300), s[0]

# both have two digits, one a multiple of the other

if mn >= 10 and mx % mn == 0:

sol.add((mn, mx))

print(s)

# output solutions

for mn, mx in sol:

print(f"{mx} play Table Tennis, a minimum of {mn} play all four.")

John Z permalink

I didn’t include my logic in comments as it’s similar to Brian’s. My code uses a more straightforward way of generating the numbers signed up for each sport.


from itertools import permutations

sports=[0] * 4  # number signed up in each of the four sports

sols=[] # list of solutions to avoid printing duplicates
for digits in permutations(range(1,10),8): # digits are different, no zeroes
    for j in range(4):
        # each sport is a different 2 digit value
        sports[j] = digits[j*2]*10 + digits[j*2+1] 

    sports.sort() # sort sports into ascending order
                  # table tennis has fewest participants
    max4 = sports[0] # max signed up for all 4 = sport with fewest signed up
    min4 = sum(sports)-300   # min signed up for all four
    if (min4 >= 10        # minimum must be a 2 digit number
        and max4%min4==0     # max is an integral multiple of min
        and tuple(sports) not in sols): # skip duplicates
        sols.append(tuple(sports)) # add solution to list; tuple() gives snapshot
        print(*sports, sep=", ")
        
print("Table tennis:",sols[0][0])

from itertools import permutations

sports=[0] * 4 # number signed up in each of the four sports

sols=[] # list of solutions to avoid printing duplicates

for digits in permutations(range(1,10),8): # digits are different, no zeroes

for j in range(4):

# each sport is a different 2 digit value

sports[j] = digits[j*2]*10 + digits[j*2+1]

sports.sort() # sort sports into ascending order

# table tennis has fewest participants

max4 = sports[0] # max signed up for all 4 = sport with fewest signed up

min4 = sum(sports)-300 # min signed up for all four

if (min4 >= 10 # minimum must be a 2 digit number

and max4%min4==0 # max is an integral multiple of min

and tuple(sports) not in sols): # skip duplicates

sols.append(tuple(sports)) # add solution to list; tuple() gives snapshot

print(*sports, sep=", ")

print("Table tennis:",sols[0][0])

Sunday Times Teaser 3022 – Sporty Set

by Victor Bryant

Published Sunday August 23 2020 (link)

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