Sunday Times Teaser 2877 – Four Steadings and a Numeral

by Stephen Hogg

Published November 12 2017 (link)

Farmers Al, Bo, Cy and Di have different double-digit numbers of sheep kept in their respective steadings. Al has the fewest and his number of sheep is a certain fraction of Bo’s number of sheep. Also, Bo’s number of sheep is that same fraction of Cy’s number, and Cy’s number is that same fraction of Di’s.

If I told you the total of Bo’s number of sheep added to Cy’s, then you would be unable to work out all their numbers of sheep. Similarly, if instead I told you just Bo’s number of sheep, then you would be unable to work out all the other numbers.

What (in the order Al, Bo, Cy, Di) are their numbers of sheep?

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Brian Gladman permalink


from fractions import Fraction as RF
from collections import defaultdict

# generate four two-digit numbers in increasing order such 
# that the ratios between each number and its successor are 
# all the same 
def generate():
  # select the largest number (D)
  for D in range(10, 100):
    # the next largest number (C) is less than D
    for C in range(10, D):
      # find the constant ratio from C and D and
      # compute A and B
      f = RF(C, D)
      B = f * C
      A = f * B
      # if A has two digits and A and B are both integers
      if A >= 10 and A.denominator == B.denominator == 1:
        yield A.numerator, B.numerator, C, D

# collect items in the sequence <seq> into groups with the
# same result when the function <f> is applied to them; 
# find for such items those for which the function <g>, 
# when applied to an item does not give a unique result
def filter(seq, f, g):
  d, e = defaultdict(list), defaultdict(set)
  for t in seq:
    k = f(t)
    d[k].append(t)
    e[k].add(g(t))
  m = set()
  for k, v in e.items():
    if len(v) > 1:
      m.update(d[k])
  return m

# collect a list of possible sequences
lt1 = list(generate())

# filter out those sequences for which the sum of B and C
# identifies a unique result for (B, C)
lt2 = filter(lt1, lambda t: t[1] + t[2], lambda t: (t[1], t[2]))

# filter out those sequences for which B identifies 
# a unique result for (A, D)
lt3 = filter(lt1, lambda t: t[1], lambda t: (t[0], t[3]))

# the solution must be in both these groups
sol, = lt2 & lt3
fs = 'Al, Bo, Cy and Di respectively own {0}, {1}, {2} and {3} sheep.'
print(fs.format(*sol))

from fractions import Fraction as RF

from collections import defaultdict

# generate four two-digit numbers in increasing order such

# that the ratios between each number and its successor are

# all the same

def generate():

# select the largest number (D)

for D in range(10, 100):

# the next largest number (C) is less than D

for C in range(10, D):

# find the constant ratio from C and D and

# compute A and B

f = RF(C, D)

B = f * C

A = f * B

# if A has two digits and A and B are both integers

if A >= 10 and A.denominator == B.denominator == 1:

yield A.numerator, B.numerator, C, D

# collect items in the sequence <seq> into groups with the

# same result when the function <f> is applied to them;

# find for such items those for which the function <g>,

# when applied to an item does not give a unique result

def filter(seq, f, g):

d, e = defaultdict(list), defaultdict(set)

for t in seq:

k = f(t)

d[k].append(t)

e[k].add(g(t))

m = set()

for k, v in e.items():

if len(v) > 1:

m.update(d[k])

return m

# collect a list of possible sequences

lt1 = list(generate())

# filter out those sequences for which the sum of B and C

# identifies a unique result for (B, C)

lt2 = filter(lt1, lambda t: t[1] + t[2], lambda t: (t[1], t[2]))

# filter out those sequences for which B identifies

# a unique result for (A, D)

lt3 = filter(lt1, lambda t: t[1], lambda t: (t[0], t[3]))

# the solution must be in both these groups

sol, = lt2 & lt3

fs = 'Al, Bo, Cy and Di respectively own {0}, {1}, {2} and {3} sheep.'

print(fs.format(*sol))

Sunday Times Teaser 2877 – Four Steadings and a Numeral

by Stephen Hogg

Published November 12 2017 (link)

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