Question
Suppose you spend $23, and you pay with a $50 bill. The cashier gives your change in bills. List all the different ways you can receive your correct change If you will not take more than 10 bills, and the cashier only has 1s, 5s, and 10s. Show your work
Answers
GPT-4o mini
To determine the different ways to receive your correct change of $27, we can set up the problem in terms of the number of 1-dollar bills (let's call it \(x\)), the number of 5-dollar bills (let's call it \(y\)), and the number of 10-dollar bills (let's call it \(z\)).
The total amount of change we want is \(x + 5y + 10z = 27\).
Additionally, the condition that we do not take more than 10 bills can be expressed as \(x + y + z \leq 10\).
Let's analyze the problem systematically:
### Step 1: Solve for different combinations
We'll solve the equation for various values of \(z\) (the number of 10-dollar bills) and find the corresponding values of \(x\) and \(y\).
1. **If \(z = 0\)**:
- The equation becomes \(x + 5y = 27\).
- Now we also need \(x + y \leq 10\).
By substituting \(x = 27 - 5y\):
\[
27 - 5y + y \leq 10 \implies 27 - 4y \leq 10 \implies 17 \leq 4y \implies y \geq \frac{17}{4} \approx 4.25
\]
Therefore, \(y\) can only take values 5, 6, …, 10 (we round up to the nearest whole number). Let's go through these:
- **\(y = 5\)**:
- \(x + 25 = 27 \implies x = 2\) → (2, 5, 0)
- **\(y = 6\)**:
- \(x + 30 = 27 \implies x = -3\) → not valid.
Thus, the only solution for \(z = 0\) is \((2, 5, 0)\).
2. **If \(z = 1\)**:
- The equation becomes \(x + 5y + 10 = 27 \implies x + 5y = 17\).
- The condition becomes \(x + y + 1 \leq 10 \implies x + y \leq 9\).
By substituting \(x = 17 - 5y\):
\[
17 - 5y + y \leq 9 \implies 17 - 4y \leq 9 \implies 8 \leq 4y \implies y \geq 2
\]
Possible values for \(y\): 2, 3, 4, 5, 6, 7, 8.
- **\(y = 2\)**:
- \(x + 10 = 17 \implies x = 7\) → (7, 2, 1)
- **\(y = 3\)**:
- \(x + 15 = 17 \implies x = 2\) → (2, 3, 1)
- **\(y = 4\)**:
- \(x + 20 = 17 \implies x = -3\) → not valid.
Solutions for \(z = 1\): \((7, 2, 1)\) and \((2, 3, 1)\).
3. **If \(z = 2\)**:
- The equation becomes \(x + 5y + 20 = 27 \implies x + 5y = 7\).
- The condition becomes \(x + y + 2 \leq 10 \implies x + y \leq 8\).
By substituting \(x = 7 - 5y\):
\[
7 - 5y + y \leq 8 \implies 7 - 4y \leq 8 \implies -1 \leq 4y \implies y \geq 0
\]
Possible values for \(y\): 0, 1, 2.
- **\(y = 0\)**:
- \(x + 0 = 7 \implies x = 7\) → (7, 0, 2)
- **\(y = 1\)**:
- \(x + 5 = 7 \implies x = 2\) → (2, 1, 2)
- **\(y = 2\)**:
- \(x + 10 = 7 \implies x = -3\) → not valid.
Solutions for \(z = 2\): \((7, 0, 2)\) and \((2, 1, 2)\).
4. **If \(z = 3\)**:
- The equation becomes \(x + 5y + 30 = 27 \implies x + 5y = -3\) → not valid.
### Step 2: Summary of Solutions
The valid combinations of \((x, y, z)\) that provide change of $27 using at most 10 bills:
- \((2, 5, 0)\) → 2 one-dollar bills, 5 five-dollar bills
- \((7, 2, 1)\) → 7 one-dollar bills, 2 five-dollar bills, 1 ten-dollar bill
- \((2, 3, 1)\) → 2 one-dollar bills, 3 five-dollar bills, 1 ten-dollar bill
- \((7, 0, 2)\) → 7 one-dollar bills, 0 five-dollar bills, 2 ten-dollar bills
- \((2, 1, 2)\) → 2 one-dollar bills, 1 five-dollar bill, 2 ten-dollar bills
### Total Number of Ways
Therefore, the total number of ways you can receive your change is **5**.
The total amount of change we want is \(x + 5y + 10z = 27\).
Additionally, the condition that we do not take more than 10 bills can be expressed as \(x + y + z \leq 10\).
Let's analyze the problem systematically:
### Step 1: Solve for different combinations
We'll solve the equation for various values of \(z\) (the number of 10-dollar bills) and find the corresponding values of \(x\) and \(y\).
1. **If \(z = 0\)**:
- The equation becomes \(x + 5y = 27\).
- Now we also need \(x + y \leq 10\).
By substituting \(x = 27 - 5y\):
\[
27 - 5y + y \leq 10 \implies 27 - 4y \leq 10 \implies 17 \leq 4y \implies y \geq \frac{17}{4} \approx 4.25
\]
Therefore, \(y\) can only take values 5, 6, …, 10 (we round up to the nearest whole number). Let's go through these:
- **\(y = 5\)**:
- \(x + 25 = 27 \implies x = 2\) → (2, 5, 0)
- **\(y = 6\)**:
- \(x + 30 = 27 \implies x = -3\) → not valid.
Thus, the only solution for \(z = 0\) is \((2, 5, 0)\).
2. **If \(z = 1\)**:
- The equation becomes \(x + 5y + 10 = 27 \implies x + 5y = 17\).
- The condition becomes \(x + y + 1 \leq 10 \implies x + y \leq 9\).
By substituting \(x = 17 - 5y\):
\[
17 - 5y + y \leq 9 \implies 17 - 4y \leq 9 \implies 8 \leq 4y \implies y \geq 2
\]
Possible values for \(y\): 2, 3, 4, 5, 6, 7, 8.
- **\(y = 2\)**:
- \(x + 10 = 17 \implies x = 7\) → (7, 2, 1)
- **\(y = 3\)**:
- \(x + 15 = 17 \implies x = 2\) → (2, 3, 1)
- **\(y = 4\)**:
- \(x + 20 = 17 \implies x = -3\) → not valid.
Solutions for \(z = 1\): \((7, 2, 1)\) and \((2, 3, 1)\).
3. **If \(z = 2\)**:
- The equation becomes \(x + 5y + 20 = 27 \implies x + 5y = 7\).
- The condition becomes \(x + y + 2 \leq 10 \implies x + y \leq 8\).
By substituting \(x = 7 - 5y\):
\[
7 - 5y + y \leq 8 \implies 7 - 4y \leq 8 \implies -1 \leq 4y \implies y \geq 0
\]
Possible values for \(y\): 0, 1, 2.
- **\(y = 0\)**:
- \(x + 0 = 7 \implies x = 7\) → (7, 0, 2)
- **\(y = 1\)**:
- \(x + 5 = 7 \implies x = 2\) → (2, 1, 2)
- **\(y = 2\)**:
- \(x + 10 = 7 \implies x = -3\) → not valid.
Solutions for \(z = 2\): \((7, 0, 2)\) and \((2, 1, 2)\).
4. **If \(z = 3\)**:
- The equation becomes \(x + 5y + 30 = 27 \implies x + 5y = -3\) → not valid.
### Step 2: Summary of Solutions
The valid combinations of \((x, y, z)\) that provide change of $27 using at most 10 bills:
- \((2, 5, 0)\) → 2 one-dollar bills, 5 five-dollar bills
- \((7, 2, 1)\) → 7 one-dollar bills, 2 five-dollar bills, 1 ten-dollar bill
- \((2, 3, 1)\) → 2 one-dollar bills, 3 five-dollar bills, 1 ten-dollar bill
- \((7, 0, 2)\) → 7 one-dollar bills, 0 five-dollar bills, 2 ten-dollar bills
- \((2, 1, 2)\) → 2 one-dollar bills, 1 five-dollar bill, 2 ten-dollar bills
### Total Number of Ways
Therefore, the total number of ways you can receive your change is **5**.