Another mensa puzzle – salman faris

My friend Zatyhan dropped me with yet another Mensa puzzle so I decided to attempt it.

This is Mensa Puzzle 877 from the The Actuary Jan/Feb 2025 edition titled Windfall win.

Windfall win. “A bank cashier making a payment to a customer transposed the pounds and pence, which gave the customer far too much money. After the customer had spent £18.85, he still had twice the amount he should have received from the cashier. How much should they have received if there hadn’t been an error?”

Now I was quite furious with how the puzzle was written since it felt confusing at first, second and even a third glance. It felt like Mensa is testing for my English skills rather than my problem solving ability.

What I settled on is that there are four parts to focus on here:

“A bank cashier making a payment to a customer transposed the pounds and pence, which gave the customer far too much money” – OK. Note the word bank cashier pays to a customer. It is also unsettling for me that transpose is not properly defined, and this will lead to problems as we shall see.
“After the customer had spent £18.85” – this money was spent elsewhere by the customer!
“he still had twice the amount he should have received from the cashier.” – cashier here is referring to the bank cashier! Yeah wth?!
“How much should they have received if there hadn’t been an error?” – he should received from the bank cashier had there not been an error.

Modelling this

So modelling this, we have:

The bank cashier is supposed to pay the customer a sum S' = 100x + p where x is the number in pounds and p in pence. Here I convert the pounds part to pence by multiplying by 100. The paid sum was done by a mistake, where the pounds and pence are transposed (i.e. swapped), so the bank cashier paid the customer an error sum of S = 100p + x instead.
The customer spends £18.85 and gets a change C = S - 18.85.
But the change is still twice the amount he should have received from the bank cashier. So he obtained 2S' when it should have been S' - 18.85.
The question is, what is S'?

Solving the problem

So we can collate the problem into a single equation.

S - 18.85 = 2S'.

The equation above can be written equivalently as

100p + x - 18.85 = 2(100x + p).

The strategy is that if we obtain x and p, we obtain S'.

We can rearrange the terms above to obtain a linear Diophantine equation

98p + (-199)x = 1885.

The Euclidean algorithm gives us \mathrm{gcd}(98, 199) = 1 which divides 1885 so this equation has integer solutions (in fact, infinitely many). That’s awesome, at least I know I’m not being pranked by Mensa.

If you actually do the hard work, you will obtain

\begin{align*} x_0 &= -33 \cdot 1885 = -62205, \\ p_0 &= -67 \cdot 1885 = -126295, \end{align*}

as initial solutions. This gives the general solution to be

\begin{align*} x &= -62205 + 98k, \\ p &= -126295 + 199k, \end{align*}

for any integer k \in \mathbb{Z}.

Now x and p have to be positive numbers since we are dealing with money. We found that both x \geqslant 0 and p \geqslant 0 are satisfied iff

k > \left\lfloor \max \left\{ \frac{62205}{98}, \frac{126295}{199} \right\} \right\rfloor = 634.

So putting k = 635, we obtain the first solution to the problem which is x = 25, p = 70 or equivalently, £25.70.

In fact, this is the unique solution to the problem… in some sense… which I only partially agree to.

Here’s the argument. If k = 636, we obtain the next solution pair of x = 123, p = 269 or equivalently, £125.69. In this case, the bank cashier mistakenly gave £270.23 to our customer and we can verify that indeed this number satisfies our problem

\underbrace{270.23}_{=S} - 18.85 = 251.38 = 2 ( \underbrace{125.69}_{=S'}).

The problem is that the mistake that the bank cashier made was transposing the pounds and the pence, and that you can argue that the act of transposing only makes sense if both the pounds and the pence are two digits. So £125.69 being the supposed change being “directly” transposed would have given £69.125. Whereas I would have appropriated the transpose to be £69 and 125 pence which gives £70.25.

So there’s this ambiguity that would have been resolved had “transpose” been properly defined. In this sense, any solution with more than two digits in x or p would have been disregarded. And it’s obvious then that only x = 25, p = 70 gives the correct solution within the problem’s (not very well-defined) constraints.