Sunday Times Teaser 2813 – Katy’s Piggy Banks

by Bob Walker

Published: 21 August 2016 (link)

Our granddaughter Katy has two piggy banks containing a mixture of 10p and 50p coins, making a grand total of £10. The first piggy bank contains less than the other, and in it the number of 50p coins is a multiple of the number of 10p coins.

Katy chose a coin at random from the first piggy bank and then put it in the second. Then she chose a coin at random from the second and put it in the first. She ended up with the same amount of money in the first piggy bank as when she started. In fact there was a 50:50 chance of this happening.

How much did she end up with in the first piggy bank?

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Brian Gladman permalink

This one is straightforward by hand and pretty well any analysis is sufficient to produce a Python solution in a few lines. So here is a solution with essentially no prior analysis:


from fractions import Fraction as RF

# the number of 50p coins in the first piggy bank
for F in range(1, 10):
  # the number of 10p coins in the first piggy bank
  for T in range(1, F):
    
    # the number of 50p coins is a multiple (M) of the
    # number of 10p ones; the total is less than £5
    M, r = divmod(F, T)
    if r or 50 * F + 10 * T >= 500:
      continue
          
    # the number of 50p coins in the second piggy bank
    for V in range(1, 20 - F):
        
        # find the number of 10p coins (U) in the second
        # piggy bank from the known overall total of £10
        U = 100 - 5 * (F + V) - T

        # now calculate the probability that the two coin
        # exchanges won't alter the amounts in the banks
        p = ( RF(1, M + 1) * RF(U + 1, U + V + 1) +
              RF(M, M + 1) * RF(V + 1, U + V + 1) )
        
        # which we are told is 50:50
        if p == RF(1, 2):
          print('\u00a3{:.2f}'.format((T + 5 * F) / 10))

from fractions import Fraction as RF

# the number of 50p coins in the first piggy bank

for F in range(1, 10):

# the number of 10p coins in the first piggy bank

for T in range(1, F):

# the number of 50p coins is a multiple (M) of the

# number of 10p ones; the total is less than £5

M, r = divmod(F, T)

if r or 50 * F + 10 * T >= 500:

continue

# the number of 50p coins in the second piggy bank

for V in range(1, 20 - F):

# find the number of 10p coins (U) in the second

# piggy bank from the known overall total of £10

U = 100 - 5 * (F + V) - T

# now calculate the probability that the two coin

# exchanges won't alter the amounts in the banks

p = ( RF(1, M + 1) * RF(U + 1, U + V + 1) +

RF(M, M + 1) * RF(V + 1, U + V + 1) )

# which we are told is 50:50

if p == RF(1, 2):

print('\u00a3{:.2f}'.format((T + 5 * F) / 10))

Let the first piggy bank start with \(T\) 10p coins and \(M T\) 50p coins; let the second piggy bank start with \(U\) 10p and \(V\) 50p coins.

The probability of choosing a 10p coin for tranfer is \(1 / (M + 1)\) and the probability a 10p coin will then be moved back is \((U + 1) /(U + V + 1)\).

The probability of choosing a 50p coin for tranfer is \(M / (M + 1)\) and the probability a 50p coin will then be moved back is \((V + 1) /(U + V + 1)\).

Hence the probability that the two exchanges will leave the piggy banks unchanged is: \[\frac{(U + 1) + M(V + 1)}{(M + 1)(U + V + 1)}\] Equating this to 1/2 now gives: \[U – V = \frac{M + 1}{M – 1}\] Since \(U\) and \(V\) are integers, this means that \(M\) can only be 2 or 3.

The total sum in both piggy banks is £10 so: \[(5M + 1)T + U + 5V = 100\]\[(5M + 1)T + 6V + (M + 1) / (M – 1) = 100\]This gives three possible solution:\[\begin{array}{cccl} M = 2: & 11T + 6V = 97 & \Rightarrow &T = 5, U = 10, V = 7 \\ M = 3: & 8T + 3V = 49 & \Rightarrow &T = 2, U = 13, V = 11 \\ & & \Rightarrow &T = 5, U = 5, V = 3 \end{array}\]

But, the value of the coins in the first piggy bank is less than £5, so \(10(5M + 1)T < 500\), which leaves only one solution: \(T = 2, U = 13\) and \(V = 11\).

So there is £3.20 in the first piggy bank and £6.80 in the second.

Sunday Times Teaser 2813 – Katy’s Piggy Banks

by Bob Walker

Published: 21 August 2016 (link)

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