## Task 10/65

A gentleman cashes a check at the bank and does not check the amount paid out. In a store, he pays a bill of 26.66 MDN from this money. To his surprise, the remaining amount is twice as large as the amount the check was issued for. The gentleman therefore goes back to the bank, where it turns out that the cashier had switched the marks and pfennigs.

What was the amount of the check?
The check was for $x$ marks and $y$ pfennigs $=(100 x+y)$ pfennigs. Paid out were $y$ marks and $x$ pfennigs $=(100 y+x)$ pfennigs. From this, 26 marks and 66 pfennigs $=2666$ pfennigs were spent. A remainder of $2 x$ marks and $2 y$ pfennigs $=(200 x+2 y)$ pfennigs remained. Thus, the equation is

$$
100 y+x-2666=200 x+2 y \rightarrow y=\frac{199 x+2666}{98}=2 x+27+\frac{3 x+20}{98}
$$

For $x$ and $y$ to be integers, there must be an integer $u$ such that $3 x+20=98 u$. Therefore,

$$
x=\frac{98 u-20}{3}=32 u-7+\frac{2 u+1}{3}
$$

For $x$ to be an integer, only the values $u=1 ; u=4 ; u=7 ; \ldots u=3 k+1$ with $k=0 ; 1 ; 2 ; \ldots$ are possible. This results in $x=26$ as the only value (since $x$ was a pfennig amount, $x \leq 99$, which is only satisfied for $u=1$). This also yields $y=80$.

Thus, the check was for 26.80 MDN, 80.26 MDN were mistakenly paid out, and after settling the bill for 26.66 MDN, 53.60 MDN remained, which is double 26.80 MDN.