Sunday Times Teaser 2635 – Three Sisters

by Nick MacKinnon

A father shared his estate among his three daughters with each getting a fraction of it.

In their simplest form, these fractions had numerators less than 100 and shared a common denominator.

The largest fraction was less than three times the smallest and each fraction recurred after three places when expressed in decimal form (e.g. 0.abcabc…) .

Each of the digits 1 to 9 occurred in one of these three decimals.

What were the three fractions?

4 Comments Leave one →

brian gladman permalink

Here is my version.


from fractions import Fraction

# if the fraction f has the form 0.ABCABC... then:
# f = ABC / 1000 + ABC / 1000^2 + .... = ABC / 999

# the first value
for ABC in range(1, 333):
  # the second value
  for DEF in range(ABC + 1, (999 - ABC) // 2):
    # the third value
    GHI = 999 - ABC - DEF
    # apply the constraints
    if DEF < GHI < 3 * ABC:
      # check that all digits in (1 .. 9) are used in the decimals
      if set(str(ABC) + str(DEF) + str(GHI)) == set('123456789'):
        # now convert to fractions
        t = tuple(Fraction(x, 999) for x in (ABC, DEF, GHI))
        if t[0].denominator == t[1].denominator == t[2].denominator:
          if all(x.numerator < 100 for x in t):
            f = ((x.numerator, x.denominator) for x in t)
            s = ' and '.join('/'.join(str(x) for x in y) for y in f)
            print('The shares are', s.replace(' and ', ', ', 1))

from fractions import Fraction

# if the fraction f has the form 0.ABCABC... then:

# f = ABC / 1000 + ABC / 1000^2 + .... = ABC / 999

# the first value

for ABC in range(1, 333):

# the second value

for DEF in range(ABC + 1, (999 - ABC) // 2):

# the third value

GHI = 999 - ABC - DEF

# apply the constraints

if DEF < GHI < 3 * ABC:

# check that all digits in (1 .. 9) are used in the decimals

if set(str(ABC) + str(DEF) + str(GHI)) == set('123456789'):

# now convert to fractions

t = tuple(Fraction(x, 999) for x in (ABC, DEF, GHI))

if t[0].denominator == t[1].denominator == t[2].denominator:

if all(x.numerator < 100 for x in t):

f = ((x.numerator, x.denominator) for x in t)

s = ' and '.join('/'.join(str(x) for x in y) for y in f)

print('The shares are', s.replace(' and ', ', ', 1))

ahmet cetinbudaklar permalink

Only one trio of this set is the answer obeying the given constraints:
5/37=0.135.. , 6/37=0.162.. , 7/37=0.189.. , 8/37=0.216.. , 9/37=0.243.. , 11/37=0.297.. , 12/37=0.324.. , 13/37=0.351.. , 14/37=0.378.. , 16/37=0.432.. , 17/37=0.459.. , 18/37=0.486.. , 19/37=0.513.. , 21/37=0.567.. , 22/37=0.594..

Reply
- brian gladman permalink
  
  HI Ahmet,
  
  Your list is longer than it needs to be.
  
  22 is not possible because this only leaves 15 for the sum of the other two. So the smaller of these has to be 7 or less and the maximum hence has to be less than 21, which it isn’t.
  
  The same is true for 21 so the maximum value is 19/ 37 (as you suggest, 20 isn’t possible).
  
  Similarly if the smallest is 5/37, the other two sum to 32/37 and this makes the maximum at least 16/37. But this is not possible because it is larger than 15 – three times the smallest.
  
  So the minimum is 6/37 and the maximum is 19/37.
  
  Reply
ahmet cetinbudaklar permalink

Hello Brian:

Thjank you for your correction by lessening of my list by 2+1 from both ends which makes life even easier.
I must say I have been enjoying Sunday Times teasers aklmost right from the beginning but I do miss the Observer “Brain Twisters” of the 1960’s by Dr. Barnard and never continued after his farewell to life.
Still this is what we have and carry on dancing.
Have a nice Easter.
ahmet

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Sunday Times Teaser 2635 – Three Sisters

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