Sunday Times Teaser 2802 – That’s the Ticket!

by Nick MacKinnon

Published: 5 June 2016 (link)

Alan, Betty and Charlie each chose a different set of six numbers (from 1 to 49) for their lottery ticket. In each case the product of the six numbers was a perfect square and also each set of six numbers used each of the digits 0 to 9 exactly once.

Alan won the lottery by getting all six numbers correct. Betty and Charlie also got prizes because they each had at least three numbers correct.

What were Alan’s six numbers?

2 Comments Leave one →

Brian Gladman permalink


from itertools import combinations, permutations
from operator import mul
from functools import reduce

# for possible sets of six numbers
sol = set()

# there are two single digit numbers and four two digit ones
# and the tens digits of the latter must be 1 .. 4 inclusive
tens = tuple(range(1, 5))

# try the other non-zero digits for the single digit numbers
for ab in combinations(range(5, 10), 2):
  
  # permute the remaining digits for the two digit numbers unit digits
  for units in permutations(set(range(10)).difference(tens + ab), 4):
        
    # form the six numbers
    nbrs = ab + tuple([10 * x + y for x, y in zip(tens, units)])
    
    # and save them if their product is a perfect square 
    pr = reduce(mul, nbrs, 1)
    if pr == int(pr ** 0.5) ** 2:
      sol.add(nbrs)

# look for a solution that shares three or more numbers with all
# other solutions
for nbrs in sol:
  if all(len(set(nbrs) & set(x)) >= 3 for x in sol if x != nbrs):
    i, _, f = (', '.join(str(x) for x in nbrs)).rpartition(', ')
    print("Alan's numbers were {}.".format(' and '.join((i, f))))

from itertools import combinations, permutations

from operator import mul

from functools import reduce

# for possible sets of six numbers

sol = set()

# there are two single digit numbers and four two digit ones

# and the tens digits of the latter must be 1 .. 4 inclusive

tens = tuple(range(1, 5))

# try the other non-zero digits for the single digit numbers

for ab in combinations(range(5, 10), 2):

# permute the remaining digits for the two digit numbers unit digits

for units in permutations(set(range(10)).difference(tens + ab), 4):

# form the six numbers

nbrs = ab + tuple([10 * x + y for x, y in zip(tens, units)])

# and save them if their product is a perfect square

pr = reduce(mul, nbrs, 1)

if pr == int(pr ** 0.5) ** 2:

sol.add(nbrs)

# look for a solution that shares three or more numbers with all

# other solutions

for nbrs in sol:

if all(len(set(nbrs) & set(x)) >= 3 for x in sol if x != nbrs):

i, _, f = (', '.join(str(x) for x in nbrs)).rpartition(', ')

print("Alan's numbers were {}.".format(' and '.join((i, f))))

Jim Randell permalink

Here’s my solution:


from itertools import permutations
from enigma import multiply, is_square, printf

# the numbers are: a, b, 1c, 2d, 3e, 4f
# where the remaining digits are 0, 5, 6, 7, 8, 9

# find possible sets of numbers
ss = list()
for (a, b, c, d, e, f)  in permutations((0, 5, 6, 7, 8, 9)):
  if a == 0 or b < a: continue
  s = (a, b, 10 + c, 20 + d, 30 + e, 40 + f)
  # we only want sets whose product is a perfect square
  if is_square(multiply(s)) is None: continue
  ss.append(s)
  printf("{s}")

# now choose a set for A
for A in ss:
  sA = set(A)
  # and find other sets that share at least 3 numbers
  BC = list(s for s in ss if s != A and len(sA.intersection(s)) > 2)
  if len(BC) > 1:
    printf("A={A}, BC={BC}")

from itertools import permutations

from enigma import multiply, is_square, printf

# the numbers are: a, b, 1c, 2d, 3e, 4f

# where the remaining digits are 0, 5, 6, 7, 8, 9

# find possible sets of numbers

ss = list()

for (a, b, c, d, e, f) in permutations((0, 5, 6, 7, 8, 9)):

if a == 0 or b < a: continue

s = (a, b, 10 + c, 20 + d, 30 + e, 40 + f)

# we only want sets whose product is a perfect square

if is_square(multiply(s)) is None: continue

ss.append(s)

printf("{s}")

# now choose a set for A

for A in ss:

sA = set(A)

# and find other sets that share at least 3 numbers

BC = list(s for s in ss if s != A and len(sA.intersection(s)) > 2)

if len(BC) > 1:

printf("A={A}, BC={BC}")

Again I’ve used some useful routines from the enigma.py library [ https://www.magwag.plus.com/jim/enigma.html ].

There are only three possible sets of six numbers that use the digits 0-9 exactly once and multiply together to give a square. Fortunately one of them shares 3 or more numbers with the other two, so this gives us A’s numbers.

Sunday Times Teaser 2802 – That’s the Ticket!

by Nick MacKinnon

Published: 5 June 2016 (link)

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