# 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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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:

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.