Sunday Times Teaser 2735 – Two to Choose

by Ian Kay

Published: 22 February 2015 (link)

I told Sue a two-figure number and I told Terry another two-figure number, one of which was a multiple of the other. I explained this to them but knew that neither of them would be able to work out the other number. (In fact, if they had to guess the other number Sue had three times as many choices as Terry — but I did not tell them that.) When Sue confirmed that it was impossible for her to work out Terry’s number, he was then able to work out her number.

What were their numbers?

2 Comments Leave one →

Brian Gladman permalink


from collections import defaultdict

# list for each number the numbers it can be paired with
n_map = defaultdict(set)
for lo in range(10, 50):
  for hi in range(2 * lo, 100, lo):
    n_map[lo].add(hi)
    n_map[hi].add(lo)

# neither Sue nor Terry can deduce the other's number so 
# their numbers must all map to more than one possibility
ambiguous = set(a for a, b_set in n_map.items() 
                      if len(b_set) > 1)

# but Terry, knowing that Sue can't deduce his number, can
# then deduce hers
pairs = []
for terry in ambiguous:
  sue_set = n_map[terry]
  # find those of Terry's numbers for which only one of
  # Sue's numbers is possible (and which can hence be
  # deduced by Terry)
  sue = sue_set & ambiguous
  if len(sue) == 1:
    pairs.append((min(sue), terry))

# we now know the possible (Sue, Terry) number pairs and can
# use the '3:1 rule' to find the pair that solves the teaaser
for sue, terry in pairs:
  if len(n_map[sue]) == 3 * len(n_map[terry]):
    print("Sue's number is {}, Terry's is {}.".format(sue, terry))

from collections import defaultdict

# list for each number the numbers it can be paired with

n_map = defaultdict(set)

for lo in range(10, 50):

for hi in range(2 * lo, 100, lo):

n_map[lo].add(hi)

n_map[hi].add(lo)

# neither Sue nor Terry can deduce the other's number so

# their numbers must all map to more than one possibility

ambiguous = set(a for a, b_set in n_map.items()

if len(b_set) > 1)

# but Terry, knowing that Sue can't deduce his number, can

# then deduce hers

pairs = []

for terry in ambiguous:

sue_set = n_map[terry]

# find those of Terry's numbers for which only one of

# Sue's numbers is possible (and which can hence be

# deduced by Terry)

sue = sue_set & ambiguous

if len(sue) == 1:

pairs.append((min(sue), terry))

# we now know the possible (Sue, Terry) number pairs and can

# use the '3:1 rule' to find the pair that solves the teaaser

for sue, terry in pairs:

if len(n_map[sue]) == 3 * len(n_map[terry]):

print("Sue's number is {}, Terry's is {}.".format(sue, terry))

Arthur Vause permalink

A broadly similar solution to Brian’s


from itertools import combinations, permutations
candidates = dict()
for a in xrange(10,100):
  othernumber = [x for x in xrange(10,100)
                if x != a and (x%a==0 or a%x==0)]
  if len(othernumber)>1:
    candidates[a] = othernumber

for terry in candidates:
  suecandidates = candidates[terry]
  if len(set(suecandidates) & set(candidates) )==1:
    sue = list(set(suecandidates) & set(candidates))[0]
    if len(candidates[sue])==3*len(candidates[terry]):
      print "Terry=",terry ",Sue= ", sue

from itertools import combinations, permutations

candidates = dict()

for a in xrange(10,100):

othernumber = [x for x in xrange(10,100)

if x != a and (x%a==0 or a%x==0)]

if len(othernumber)>1:

candidates[a] = othernumber

for terry in candidates:

suecandidates = candidates[terry]

if len(set(suecandidates) & set(candidates) )==1:

sue = list(set(suecandidates) & set(candidates))[0]

if len(candidates[sue])==3*len(candidates[terry]):

print "Terry=",terry ",Sue= ", sue

Sunday Times Teaser 2735 – Two to Choose

by Ian Kay

Published: 22 February 2015 (link)

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