Sunday Times Teaser 2502 – No Title

by H Bradley and C Higgins

Published September 5 2010 (link)

John and Pat were swimming at the pool. John started at the deep end at the same time as Pat started at the shallow end, each swimming lengths at their own steady speed. They passed each other on their first lengths and then passed each other again on their second lengths. It turned out that the length of the pool was a multiple of the distance between those first two passing points. John got out of the water after completing 48 lengths. At that moment Pat had also completed a whole number of lengths.

How many?

2 Comments Leave one →

Brian Gladman permalink


from fractions import Fraction as RF
from itertools import count

#       0 ----------x----------------y------ l
#    Deep ================================== Shallow
# John >> ----------x-----------------------
# speed j                            y------
#         ----------x----------------------- <<< Pat                    
#         ---------------------------y       speed p
# 
# Let r be the speed ratio be p / j; Pat completes N
# lengths when John completes 48 so r = N / 48.  Now
# equate the times for Pat and John to reach x and y: 
#
#   x / j = (l - x) / p ; (2.l - y) / j = (l + y) / p
#
#   (x/l) = 1 / (r + 1) ; (y/l) = (2.r - 1) / (r + 1)
#
#   (y - x) / l = 2.(r - 1) / (r + 1)
#
# Let the pool length (l) be equal to K.(y - x), with
# K an integer. So:
#
#   2.K = (r + 1) / (r - 1) = (N + 48) / (N - 48) 
#
#   N = 48 + 96 / (2.K - 1) 
#
# Since K and N are integers, 2.K - 1 must be an odd
# divisor of 96 (2^5.3), which means that K = -1 and 
# N = 16 or K = 2 and N = 80; the N = 16 solution is
# not valid since John would pass Pat for the second
# time while Pat was still on their first length, not
# their second as stated.

for N in count(49):
  K, r = divmod(N + 48, 2 * (N - 48))
  if not r:
    x = RF(48, N + 48)
    y = 2 - 3 * x
    print(f'Pat swims {N} lengths (x/l = {x}, y/l = {y}, (y - x)/l = {y - x}).')
    break

from fractions import Fraction as RF

from itertools import count

# 0 ----------x----------------y------ l

# Deep ================================== Shallow

# John >> ----------x-----------------------

# speed j y------

# ----------x----------------------- <<< Pat

# ---------------------------y speed p

# Let r be the speed ratio be p / j; Pat completes N

# lengths when John completes 48 so r = N / 48. Now

# equate the times for Pat and John to reach x and y:

# x / j = (l - x) / p ; (2.l - y) / j = (l + y) / p

# (x/l) = 1 / (r + 1) ; (y/l) = (2.r - 1) / (r + 1)

# (y - x) / l = 2.(r - 1) / (r + 1)

# Let the pool length (l) be equal to K.(y - x), with

# K an integer. So:

# 2.K = (r + 1) / (r - 1) = (N + 48) / (N - 48)

# N = 48 + 96 / (2.K - 1)

# Since K and N are integers, 2.K - 1 must be an odd

# divisor of 96 (2^5.3), which means that K = -1 and

# N = 16 or K = 2 and N = 80; the N = 16 solution is

# not valid since John would pass Pat for the second

# time while Pat was still on their first length, not

# their second as stated.

for N in count(49):

K, r = divmod(N + 48, 2 * (N - 48))

if not r:

x = RF(48, N + 48)

y = 2 - 3 * x

print(f'Pat swims {N} lengths (x/l = {x}, y/l = {y}, (y - x)/l = {y - x}).')

break

Frits permalink

The formula at line 37 forces N to be a multiple of 16.

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Sunday Times Teaser 2502 – No Title

by H Bradley and C Higgins

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