2. In a toy store, at the beginning of the day, there are 20 sticker sheets (0.30 euros each), 18 footballs (3 euros each), 5 teddy bears (5 euros each), and 8 water guns (15 euros each) in stock. The new cashier makes a mess of the administration and reports only the total amount of 75.80 euros at the end of the day for the sales of these four items, and that fewer footballs were sold than any other item. How many teddy bears were sold?
A) 1
B) 2
C) 3
D) 4
E) 5
A2. D) 4 Because three of the four items have a price in whole euros, the sticker sheets must have either 6 sold (for 1.80 euros) or 16 (for 4.80 euros). More is not possible, as there were only 20 in stock. For the other three items, there is thus 74 or 71 euros left. First, assume that there is 74 euros left. Since the teddy bears and water guns cost a multiple of 5 euros, there must have been 3 soccer balls sold (with 65 euros left) or an additional multiple of 5. But 8 soccer balls is not possible, because then more than 8 teddy bears would have to be sold, and there were not that many in stock. So there are 3. There must be at least 4 water guns sold and at least 4 teddy bears, but that is more than 65 euros together. So this is not possible.

Now assume that there are 71 euros left. According to the same reasoning, 2 soccer balls must have been sold or at least 7, but the latter is not possible again. So there is still 65 euros left for the teddy bears and water guns. Of these, at least 3 must have been sold, for at least $3 \cdot 5 + 3 \cdot 15 = 60$ euros. Therefore, there must have been exactly 3 water guns sold and to complete the amount, exactly 4 teddy bears.