On planet Enigma, the residents use a currency called the confusion. There are only two confusion bills on Enigma, one worth $7$ confusions and the other worth $12$ confusions. There are also some coins of smaller value, but each weighs over $10$ kilograms, so they are difficult to carry around.

Question

In how many ways can a resident of Enigma use only bills to purchase a toaster that costs 19 confusions?

(Everybody knows, of course, that vendors on Enigma do not give change, so residents must make their purchases with exact change.)

Answer 1

To determine how many ways the residents of Enigma can use $7$ confusion and $12$ confusion bills to purchase a toaster that costs $19$ confusions, we describe each possible combination of these bills using equations and combinations.

Let \( x \) represent the number of $7$ confusion bills, and \( y \) the number of $12$ confusion bills. We need to find integer solutions to the equation:

\[
7x + 12y = 19
\]

First, observe that since \( 19 \) is relatively small, we can systematically substitute integer values for \( x \) and check if \( y \) is non-negative and an integer.

### Case \( x = 0 \):
\[
7(0) + 12y = 19 \implies 12y = 19 \implies y = \frac{19}{12} \not\in \mathbb{Z}
\]
Thus, \( y \) is not an integer, so no solution here.

### Case \( x = 1 \):
\[
7(1) + 12y = 19 \implies 7 + 12y = 19 \implies 12y = 12 \implies y = 1
\]
This gives the solution \( (1, 1) \).

### Case \( x = 2 \):
\[
7(2) + 12y = 19 \implies 14 + 12y = 19 \implies 12y = 5 \implies y = \frac{5}{12} \not\in \mathbb{Z}
\]
Again, \( y \) is not an integer, so no solution here.

### Case \( x = 3 \):
\[
7(3) + 12y = 19 \implies 21 + 12y = 19 \implies 12y = -2 \implies y = \frac{-2}{12} \not\in \mathbb{Z}
\]
Here, \( y \) is not an integer either, so no solution.

### Case \( x = 4 \):
\[
7(4) + 12y = 19 \implies 28 + 12y = 19 \implies 12y = -9 \implies y = \frac{-9}{12} = -\frac{3}{4} \not\in \mathbb{Z}
\]
\( y \) is once again not an integer.

For values of \( x \geq 5 \):
\[
7x \geq 35
\]
Since any value of \( 7x \) with \( x \geq 5 \) already exceeds the total cost of $19$ confusions, further checking is unnecessary.

By inspecting all the cases, the only valid solution occurs at \( (x, y) = (1, 1) \), which corresponds to using one $7$ confusion bill and one $12$ confusion bill.

Hence, there is precisely one way to use the $7$ and $12$ confusion bills to purchase a toaster costing $19$ confusions. Finally, we confirm that:

\[
\boxed{1}
\]