Question
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.
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.)
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.)
Answers
GPT-4o
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:
\[
\boxed{1}
\]
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:
\[
\boxed{1}
\]