###### tags: `MATA02-2022`
# Ungraded problems 4

You are to work on these questions only for the benefit of learning the material. These problems are meant for practice, not an assignment to be handed in.

## 1.
For any equation $mx + ny = 0$, where $m$ and $n$ are given integers, there is an integer solution by setting $x=0$ and $y=0$. There is also another integer solution by setting $x=n$ and $y=-m$.

For example, the equation $100x + 30y=0$ has integer solutions ($x=0, y=0$) and ($x=30, y=-100$).

But in fact, there are infinitely many solutions, which you can get by dividing the entire equation by the greatest common divisor of $m$ and $n$, and then taking all multiples of that pair as the values of $y$ and $x$ respectvely.

For example, starting from $100x + 30y = 0$, we divide by 10, giving $10x + 3y = 0$. Then any multiple of $(x=3, y=-10)$, such as $(x=-6, y =20)$ or $(x=9, y =-30)$ is another solution to the original equation.

Find all integer solutions to the following equations:

1. $105x + 24y = 0$
2. $1015x + 231y = 0$

## 2.
Determine if each of the following equations has integer solutions $x$ and $y$. Justify your answer. (Hint: remember that you know how to make the greatest common divisor using combinations)

Example: $100x + 30y = 5$ does not have any integer solutions. However, $100x + 30y = 20$ does have integer solutions, and you can justify this using the greatest comon divisor of 100 and 30.

1. $105x + 24y = 3$
2. $105x + 24y = 4$
3. $105x + 24y = 9$
4. $1015x + 231y = 9$
5. $1015x + 231y = 28$

## 3.
For each case of the five cases in problem 2 in which there is an integer solution, find two different integer solutions.
**Hint:** You can find one integer solution by working backwards using the Euclidean algorithm. Use Problem 1 to find two other solutions; remember that you can always add by 0.

**Example:**
- one solution to $100x + 30y=20$ is $x = 2, y = -6$
- one solution to $100x + 30y = 0$ is $x=3, y=-10$
- Thus, $x= 2 + 3 = 5, y = -6-10=-16$ is another solution to $100x + 30y = 20$.
- Alternately, $x = 2+3\cdot 10=32, y = -6-10\cdot 10 = -106$ is yet another solution.

## 4.
Use Eratosthene's Sieve to find all the primes between 235 and 265.

# Selected solutions
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