## Task 1 - 190831

Klaus tells:

"When I was shopping recently, I had exactly three coins with me. When paying, I noticed the following. If I hand over two of my coins, I still lack $3.50 \mathrm{M}$ to the full price of the purchased item, but if I only lay down the remaining coin, I get $3.50 \mathrm{M}$ back."

Determine from these facts all possibilities for how many coins of what kind Klaus had with him! Only 1, 5, 10, 20, and 50 Pfennig as well as 1, 2, 5, 10, and 20 Mark are to be considered.
The difference between the sum of the values of the first two coins and the value of the third coin is exactly $7.00 \mathrm{M}$. Therefore, the value of the third coin is greater than $7 \mathrm{M}$, so it must be $10 \mathrm{M}$ or $20 \mathrm{M}$. If the third coin had been a $20 \mathrm{M}$ coin, then the sum of the values of the other two coins would have had to be exactly 13 marks, which is not possible with the given coin values.

Therefore, the third coin was a $10 \mathrm{M}$ coin, and the sum of the values of the other two coins was exactly $3 \mathrm{M}$. This is only possible with the given coin denominations if Klaus had a 2-mark coin and a 1-mark coin.

So, Klaus had a 10-mark coin, a 2-mark coin, and a 1-mark coin for this purchase.