## Task 3 - 070713

List all amounts of money up to 1 MDN that can be paid using only one-penny, five-penny, and ten-penny coins (where at least one coin of each type must be used) and for which the amount of money in pennies is exactly twice the number of coins required!
If the number of one-penny coins is denoted by $x$, the number of five-penny coins by $y$, and the number of ten-penny coins by $z$, then:

$$
x+5 y+10 z=2(x+y+z)
$$

from which we get $8 z+3 y=x$.

| $z$ | $x=3 y+8 z$ | $y$ | Amount of money (in pennies) |
| :---: | :---: | :---: | :---: |
| 1 | 11 | 1 | 26 |
|  | 14 | 2 | 34 |
|  | 17 | 3 | 42 |
|  | 20 | 4 | 50 |
|  | 23 | 5 | 58 |
|  | 26 | 6 | 66 |
|  | 29 | 7 | 74 |
|  | 32 | 8 | 82 |
|  | 35 | 9 | 90 |
|  | 38 | 10 | 98 |
| 2 | 19 | 1 | 44 |
|  | 22 | 2 | 52 |
|  | 25 | 3 | 60 |
|  | 28 | 4 | 68 |
|  | 31 | 5 | 76 |
|  | 34 | 6 | 84 |
|  | 37 | 7 | 92 |
|  | 40 | 8 | 100 |
| 3 | 27 | 1 | 62 |
|  | 30 | 2 | 70 |
|  | 33 | 3 | 78 |
|  | 36 | 4 | 86 |
|  | 39 | 5 | 94 |
| 4 | 35 | 1 | 80 |
|  | 38 | 2 | 88 |
|  | 41 | 3 | 96 |
| 5 | 43 | 1 | 98 |

Since for all $z \geq 6$ no solutions exist under the conditions of the problem, there are thus exactly 26 such amounts of money, of which exactly 25 can be paid out in only one way, and exactly one amount of money (98 Pf) can be paid out in exactly two ways according to the mentioned conditions.