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Diophantine Equations

Practise finding integer solutions to equations with more than one unknown.
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This is level 2: equations derived from real life situations. You will be awarded a trophy if you get at least 7 answers correct and you do this activity online.

Find a set of positive whole number solutions to the following equations, though there may be more.

1

A gardener is planting two types of trees: apple trees and cherry trees. If the total number of trees is 15, how many of each type (\(a\) for apple and \(c\) for cherry) could there be?
🌳

\(a=\) \( \quad c=\) ☐ ✓ ✗

2

A farmer has a number of chickens (\(c)\) and a number of sheep (\(s\)). If the total number of legs is twenty four, how many of each type of animal could he have?
🐤 🐑

\(c=\) \( \quad s=\) ☐ ✓ ✗

3

A shopkeeper sells pencils
(\(p\)) in packs of 4 and erasers (\(e\)) in packs of 3. If a customer buys a total of 27 items, how many packs of each could they have bought?
🖍️

\(p=\) \( \quad e=\) ☐ ✓ ✗

4

Maria has a collection of 5p and 10p coins. If she has a total of £1.20 in her collection, how many of each type of coin (\(f\) for 5p and \(t\) for 10p) could she have?

\(f=\) \( \quad t=\) ☐ ✓ ✗

5

There are two types of pizzas at a party: cheese pizzas (\(c\)) with 8 slices each and pepperoni pizzas (\(p\)) with 6 slices each. If there are a whole number of pizzas altogether with a total of 38 slices, how many of each type of pizza could there be?

\(c=\) \( \quad p=\) ☐ ✓ ✗

6

In a garden, there are roses (\(r\)) and tulips (\(t\)). If the difference between the number of roses and tulips is 75, how many of each type of flower could there be if there are more roses than tulips?

\(r=\) \( \quad t=\) ☐ ✓ ✗

7

A baker is making 80g cupcakes (\(c\)) and 120g muffins (\(m\)). If the box of cupcakes is 200g heavier than the box of muffins, how many of each were made?
🧁

\(c=\) \( \quad m=\) ☐ ✓ ✗

8

At a sports event, there are two types of tickets: adult tickets (\(a\)) costing £20 each and child tickets (\(c\)) costing £15 each. At the end of the event the money taken in from the sale of adult tickets was £100 more than that from child tickets. Do the maths!

\(a=\) \( \quad c=\) ☐ ✓ ✗

9

Ximena multiplies her age by 9. Yusuf multiplies his age by 11. The product Ximena gets is 99 more than Yusuf's product. How old are Ximena (\(x\)) and Yusuf (\(y\))?
✖️

\(x=\) \( \quad y=\) ☐ ✓ ✗

Check

This is Diophantine Equations level 2. You can also try:
Level 1 Level 3 Level 4

Instructions

Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.

When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.
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"A really useful set of resources - thanks. Is the collection available on CD? Are solutions available?"

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Description of Levels

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Level 1 - Linear Diophantine equations of the form \(ax+by=c\)

Level 2 - Equations derived from real life situations

Level 3 - Similar to Level 1 but with larger numbers

Level 4 - Non linear Diophantine equations

More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.

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Example

A classic example of a Diophantine equation is \(3x + 2y = 11\).
The goal is to find values for \(x\) and \(y\) that are whole numbers (integers) and satisfy this equation.

To solve this, we can start by trying different values for \(x\) and see if we can find a corresponding value for \(y\) that makes the equation true. For example, let's try \(x = 1\):

\(3(1) + 2y = 11\)
\(3 + 2y = 11\)
\(2y = 11 - 3\)
\(2y = 8\)
\(y = \frac{8}{2}\)
\(y = 4\)

So, when \(x = 1\), \(y = 4\). Therefore, one solution to the equation is \(x = 1\) and \(y = 4\).

We can check if there are other solutions by trying different values for \(x\). If we try \(x = 2\):

\(3(2) + 2y = 11\)
\(6 + 2y = 11\)
\(2y = 11 - 6\)
\(2y = 5\)

This doesn't give us an integer solution for \(y\), since 5 divided by 2 is 2.5, not a whole number. Therefore, \(x = 2\) doesn't work.

Similarly, if we try \(x = 3\):

\(3(3) + 2y = 11\)
\(9 + 2y = 11\)
\(2y = 11 - 9\)
\(2y = 2\)
\(y = \frac{2}{2}\)
\(y = 1\)

So, another solution is \(x = 3\) and \(y = 1\).

Therefore, the Diophantine equation \(3x + 2y = 11\) has two positive integer solutions:
\((x, y) = (1, 4)\) and \((x, y) = (3, 1)\).

Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.

Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.

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