Question

Suzy purchased at least one pen priced at 13 each and at least one notebook priced at 19 each. If the total price of the items purchased is $58, what is the total number of pens and notebooks purchased by Suzy? 
 

Hint:

For linear Diophantine equations of the form \(ax + by = c\), you can use modular arithmetic or simply look at the units digit to quickly eliminate possibilities. For \(13p + 19n = 58\), the units digit of \(13p\) plus the units digit of \(19n\) must sum to a number ending in 8. For \(n=1\), \(19n\) ends in 9. We need \(13p\) to end in 9 (since 9+9=18). Trying \(p=3\), \(13 \times 3 = 39\), which works.

Answer 1

Step 1: Understanding the Concept:
This problem involves solving a linear Diophantine equation, which is an equation where we are only interested in integer solutions. We are given the cost of two items and a total cost, and we need to find the number of each item purchased. The constraints are that the number of items must be positive integers.
Step 2: Key Formula or Approach:
1. Let \(p\) be the number of pens and \(n\) be the number of notebooks.
2. Set up an equation based on the total cost: \(13p + 19n = 58\).
3. Use the given constraints: \(p \ge 1\) and \(n \ge 1\), and both \(p\) and \(n\) must be integers.
4. Solve the equation by testing integer values for one of the variables. It's often easier to test values for the variable with the larger coefficient.
5. Find the unique integer pair \((p, n)\) that satisfies the equation and constraints.
6. Calculate the total number of items, which is \(p + n\).
Step 3: Detailed Explanation:
Let \(p\) be the number of pens purchased and \(n\) be the number of notebooks purchased.
The equation representing the total cost is:
\[ 13p + 19n = 58 \] We are given that Suzy purchased at least one of each, so \(p \ge 1\) and \(n \ge 1\).
We can solve this by testing values for \(n\).
If n = 1:
\[ 13p + 19(1) = 58 \] \[ 13p + 19 = 58 \] \[ 13p = 58 - 19 \] \[ 13p = 39 \] \[ p = \frac{39}{13} = 3 \] This gives us a valid solution: \(p=3\) and \(n=1\). Both are integers greater than or equal to 1.
If n = 2:
\[ 13p + 19(2) = 58 \] \[ 13p + 38 = 58 \] \[ 13p = 20 \] \[ p = \frac{20}{13} \] This is not an integer, so it is not a valid solution.
If n = 3:
\[ 13p + 19(3) = 58 \] \[ 13p + 57 = 58 \] \[ 13p = 1 \] \[ p = \frac{1}{13} \] This is not an integer.
If \(n\) is greater than 3, the value of \(19n\) will be greater than 58, which would require \(p\) to be negative. Thus, no other solutions are possible.
The only valid solution is \(p=3\) and \(n=1\).
The total number of items purchased is the sum of the number of pens and notebooks.
Total items = \(p + n = 3 + 1 = 4\).
Step 4: Final Answer:
The total number of pens and notebooks purchased by Suzy is 4.