Denomination problem

Question

Suppose that the only denominations of the available bills are 11 and 19 dollars. In a convenience store, what amounts can be paid? Justify your solution in full detail.

Answer 1

"Full detail" is context-dependent. It is not clear what it means unless one knows what has been done in your course, and what tools are accessible.

From general theory, it can be shown that all numbers $\ge (11-1)(19-1)$ are representable, and that $(11-1)(19-1) -1=179$ is not.

You can do "full detail" by brute force, perhaps computer aided. Once you find out by computation that the $11$ numbers from $180$ to $190$ are representable, then you know that all numbers above that are representable, by adding a suitable number of $11$ dollar bills to the representation of some number between $180$ and $190$. For any number greater than $190$ is congruent modulo $11$ to some number between $180$ and $190$.

More simply put, once you get $11$ in a row, you know everything is all right from then on.

For numbers between $0$ and $179$, there is useful bu complicated general theory. However, careful book keeping will do the job. It is not much fun. Of course, $0$ is representable, though the convenience store owner may not be pleased. Then $1$ to $10$ are not, $11$ is, $12$ to $18$ are not, and so on. As our numbers get larger, the probability they are representable increases, and one has to be very careful. There will be $90$ representable numbers up to $180$. Quite a lot of work