Sunday Times Teaser 2541 – Lotto Luck

by H Bradley and C Higgins

Published: 5 June 2011 (link)

Chris was lucky in the lotto, winning a whole number of pounds (under £5,000), which he shared with his five children. John got £1 more than a fifth of the total; Ciaran got £1 more than a fifth of the remainder. After Ciaran got his share, Fergal got £1 more than a fifth of the remainder, and after Fergal got his share, Brendan got £1 more than a fifth of what was left. After Brendan got his share, Chris gave Grainne £1 more than a fifth of what was left and kept the remainder (a whole number of pounds).

How much did Chris keep?

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Brian Gladman permalink


from itertools import count
from fractions import Fraction as RF

# It is faster to work the sequence in reverse from what
# is left back to the original win.

# convert a fraction to an integer but only if it is one
fn = lambda x: x if x.denominator != 1 else x.numerator

# try values for the final remainder
for rem in count():
  # a list for the intermediate remainders and the win
  r = [rem]
  
  # now run backwards through the five gifts, not
  # assuming that intermediate results are integer
  for g in range(5):
    r.append(RF(5 * (r[-1] + 1), 4))
    
  # the final (win) must be an integer
  if r[-1].denominator == 1:
    
    # work out the gifted amounts
    r = r[::-1]
    gifts = map(fn, (a - b for a, b in zip(r, r[1:])))
    
    f, s, l = ', \xa3'.join(str(x) for x in gifts).rpartition(', ')
    gs = '\xa3{} and {}'.format(f, l)
    fs = 'He kept \xa3{} from a \xa3{} win with gifts of {}.'      
    print(fs.format(rem, r[0], gs))
    break

from itertools import count

from fractions import Fraction as RF

# It is faster to work the sequence in reverse from what

# is left back to the original win.

# convert a fraction to an integer but only if it is one

fn = lambda x: x if x.denominator != 1 else x.numerator

# try values for the final remainder

for rem in count():

# a list for the intermediate remainders and the win

r = [rem]

# now run backwards through the five gifts, not

# assuming that intermediate results are integer

for g in range(5):

r.append(RF(5 * (r[-1] + 1), 4))

# the final (win) must be an integer

if r[-1].denominator == 1:

# work out the gifted amounts

r = r[::-1]

gifts = map(fn, (a - b for a, b in zip(r, r[1:])))

f, s, l = ', \xa3'.join(str(x) for x in gifts).rpartition(', ')

gs = '\xa3{} and {}'.format(f, l)

fs = 'He kept \xa3{} from a \xa3{} win with gifts of {}.'

print(fs.format(rem, r[0], gs))

break

Sunday Times Teaser 2541 – Lotto Luck

by H Bradley and C Higgins

Published: 5 June 2011 (link)

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