Sunday Times Teaser 2530

by H Bradley and C Higgins

Published: 20 March 2011 (link)

At the St Patrick’s Day races, Pat backed losers in each of the first three. Starting with £100, he bet a whole-number percentage of his money in the first race and a whole-number percentage of the remainder (but not a whole number of pounds) in the second. In the third, he bet a whole-number percentage of what was left, leaving a whole-number percentage of his £100. This final percentage was the sum of the three percentages he had bet and lost.

How much was lost on the third race?

2 Comments Leave one →

Brian Gladman permalink


# Let the percentages NOT bet at each stage be p1, p2 and p3 and
# let the remainders after each stage be r1, r2 and r3 with the
# initial amount r0. Then r1 = r0.p1 / 100, r2 = r1.p2 / 100 and
# r3 = r2.p3 / 100 giving:
#
#    p1.p2.p3 = 10^6.r3/r0
#
# Also:
#
#   100.r3/r0 = (100 - p1) + (100 - p2) + (100 - p3)
#
# Eliminating r3/r0 now allows p3 to be derived from p1 and p2:
#
#   p3 = (300 - p1 - p2).10^4 / (10^4 + p1.p2) 

from itertools import product

# the percentages NOT bet on the first and second races
for p1, p2 in product(range(1, 101), repeat=2):

  # the second bet is not a whole number of £    
  second_bet, rem = divmod(p1 * (100 - p2), 100)
  if rem:
    
    # now compute the percentage for the third race,
    # which must be a whole number
    p3, rem = divmod(10000 * (300 - p1 - p2), p1 * p2 + 10000)
    if rem == 0 and 0 <= p3 <= 100:
      
      # compute the third bet
      third_bet = p1 * p2 * (100 - p3) / 10000
      print('He lost £{} on the third race.'.format(third_bet))

# Let the percentages NOT bet at each stage be p1, p2 and p3 and

# let the remainders after each stage be r1, r2 and r3 with the

# initial amount r0. Then r1 = r0.p1 / 100, r2 = r1.p2 / 100 and

# r3 = r2.p3 / 100 giving:

# p1.p2.p3 = 10^6.r3/r0

# Also:

# 100.r3/r0 = (100 - p1) + (100 - p2) + (100 - p3)

# Eliminating r3/r0 now allows p3 to be derived from p1 and p2:

# p3 = (300 - p1 - p2).10^4 / (10^4 + p1.p2)

from itertools import product

# the percentages NOT bet on the first and second races

for p1, p2 in product(range(1, 101), repeat=2):

# the second bet is not a whole number of £

second_bet, rem = divmod(p1 * (100 - p2), 100)

if rem:

# now compute the percentage for the third race,

# which must be a whole number

p3, rem = divmod(10000 * (300 - p1 - p2), p1 * p2 + 10000)

if rem == 0 and 0 <= p3 <= 100:

# compute the third bet

third_bet = p1 * p2 * (100 - p3) / 10000

print('He lost £{} on the third race.'.format(third_bet))

Frits permalink

I wanted to publish a program but rearranging the analysis above lead to a manual solution.


# Let the percentages NOT bet at each stage be p1, p2 and p3 and
# let the remainders after each stage be r1, r2 and r3 with the
# initial amount r0. Then r1 = r0.p1 / 100, r2 = r1.p2 / 100 and
# r3 = r2.p3 / 100 giving:
#
# p1 = 100.r1/r0 = 10000.r2/p2.r0 = 1000000.r3/p2.p3.r0
#
#    p1.p2.p3 = 10^6.r3/r0
#
# Also:
#
#   100.r3/r0 = (100 - p1) + (100 - p2) + (100 - p3)
#
# Eliminating r3/r0 now allows p3 to be derived from p1 and p2:
#
#   p3 = (300 - p1 - p2).10^4 / (10^4 + p1.p2)
#
#   p1 * p2 * p3 - 10^4 * (300 - p1 - p2 - p3) = 0
#
#   p1 * p2 * p3 = (300 - p1 - p2 - p3) * 16 * 5^4
#
# if p1, p2 and p3 are all multiples of 5 then
# (300 - p1 - p2 - p3) is also a multiple of 5 so
# p1 * p2 * p3 must be a multiple of 5^5
#
# so we can conclude at least 2 out (p1, p2, p3) must be equal to 75,
# remaining percentage p* must be a multiple of 16
# and (300 - p1 - p2 - p3) must be a multiple of 9 (3 * 3)
#     (150 - p*) must be a multiple of 9
#     so p* mod 9 = 6
# the only multiple of 16 and mod 9 = 6 out of 64, 80 and 96 is 96
# so (p1, p2, p3) = (75, 75, 96) in any order
#
# p1 can't be 96 as 0.75 * 96 is a whole number
# p2 can't be 96 as 75 * 0.96 is the same whole number
# so (p1, p2, p3) = (75, 75, 96) in this order
#
# on the third race was lost: (3/4)^2 * 100 * (100 - 96) / 100 = 2.25 pounds

# Let the percentages NOT bet at each stage be p1, p2 and p3 and

# let the remainders after each stage be r1, r2 and r3 with the

# initial amount r0. Then r1 = r0.p1 / 100, r2 = r1.p2 / 100 and

# r3 = r2.p3 / 100 giving:

# p1 = 100.r1/r0 = 10000.r2/p2.r0 = 1000000.r3/p2.p3.r0

# p1.p2.p3 = 10^6.r3/r0

# Also:

# 100.r3/r0 = (100 - p1) + (100 - p2) + (100 - p3)

# Eliminating r3/r0 now allows p3 to be derived from p1 and p2:

# p3 = (300 - p1 - p2).10^4 / (10^4 + p1.p2)

# p1 * p2 * p3 - 10^4 * (300 - p1 - p2 - p3) = 0

# p1 * p2 * p3 = (300 - p1 - p2 - p3) * 16 * 5^4

# if p1, p2 and p3 are all multiples of 5 then

# (300 - p1 - p2 - p3) is also a multiple of 5 so

# p1 * p2 * p3 must be a multiple of 5^5

# so we can conclude at least 2 out (p1, p2, p3) must be equal to 75,

# remaining percentage p* must be a multiple of 16

# and (300 - p1 - p2 - p3) must be a multiple of 9 (3 * 3)

# (150 - p*) must be a multiple of 9

# so p* mod 9 = 6

# the only multiple of 16 and mod 9 = 6 out of 64, 80 and 96 is 96

# so (p1, p2, p3) = (75, 75, 96) in any order

# p1 can't be 96 as 0.75 * 96 is a whole number

# p2 can't be 96 as 75 * 0.96 is the same whole number

# so (p1, p2, p3) = (75, 75, 96) in this order

# on the third race was lost: (3/4)^2 * 100 * (100 - 96) / 100 = 2.25 pounds

Sunday Times Teaser 2530

by H Bradley and C Higgins

Published: 20 March 2011 (link)

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