Solution to: Buying Bitterballs
Every natural number is member of one of the following six series:
- 0, 6, 12, 18, ...
- 1, 7, 13, 19, ...
- 2, 8, 14, 20, ...
- 3, 9, 15, 21, ...
- 4, 10, 16, 22, ...
- 5, 11, 17, 23, ...
If for a number in one of these series holds that it can be made using the numbers 6, 9, and 20, then this also holds for all subsequent numbers in the series (by adding a multiple of 6).
To find out what the largest number is that cannot be made using the numbers 6, 9, and 20, we therefore only need to know, for every series, what the smallest number is that can be made in that way.
In the series 0, 6, 12, 18, ... the smallest number that can be made is 0,
so there is no number that cannot be made.
In the series 1, 7, 13, 19, ... the smallest number that can be made is 49 (20+20+9),
so 43 is the largest number that cannot be made.
In the series 2, 8, 14, 20, ... the smallest number that can be made is 20,
so 14 is the largest number that cannot be made.
In the series 3, 9, 15, 21, ... the smallest number that can be made is 9,
so 3 is the largest number that cannot be made.
In the series 4, 10, 16, 22, ... the smallest number that can be made is 40 (20+20),
so 34 is the largest number that cannot be made.
In the series 5, 11, 17, 23, ... the smallest number that can be made is 29 (20+9),
so 23 is the largest number that cannot be made.
Therefore, 43 is the largest number that cannot be made using the numbers 6, 9, and 20.
Back to the puzzle