Let's tackle the questions one by one:

Question 7

Find the minimum positive value of $765m + 357n$ attainable by choosing suitable integers $m$ and $n$.

To find the minimum positive value of $765m + 357n$ for integers $m$ and $n$, we need to find the greatest common divisor (GCD) of 765 and 357. The minimum positive value is equal to the GCD of these numbers.

1. $765 = 3 \times 255$
2. $357 = 3 \times 119$

Now, let's find the GCD of 255 and 119 using the Euclidean algorithm:

• $255 \div 119$ gives quotient 2 and remainder 17.
• $119 \div 17$ gives quotient 7 and remainder 0.

So, the GCD of 255 and 119 is 17. Therefore, the GCD of 765 and 357 is $3 \times 17 = 51$.

The minimum positive value of $765m + 357n$ is 51.

Question 8

If $a, b, x,$ and $y$ are integers and $ax + by = 1$, then $\gcd(a, b) = \gcd(a, y) = \gcd(x, b) = \gcd(x, y) = 1$.

To address this question, we need to recognize a few facts about the greatest common divisor (GCD):

If $ax + by = 1$, then $\gcd(a, b)$ must be 1. This is because the linear combination of $a$ and $b$ resulting in 1 implies that $a$ and $b$ are coprime (their GCD is 1).

However, the statement goes further to claim that $\gcd(a, y) = \gcd(x, b) = \gcd(x, y) = 1$, which does not necessarily follow from $ax + by = 1$. Therefore, the statement is False.

Question 9

Since $x^2 - 5x + 6 = (x - 2)(x - 3)$, there is no integer $x$ for which $x^2 - 5x + 6$ is prime.

For $x^2 - 5x + 6$ to be prime, it must have only two positive divisors: 1 and itself. Factoring the quadratic equation, we get:

$x^2 - 5x + 6 = (x - 2)(x - 3)$

For any integer $x$, $(x - 2)$ and $(x - 3)$ are also integers. Since a product of two integers can only be prime if one of them is 1 and the other is the prime number itself, we check:

• If $x - 2 = 1$, then $x = 3$, and $x - 3 = 0$, not a prime.
• If $x - 3 = 1$, then $x = 4$, and $x - 2 = 2$, but 2 is not prime because $2 \times 1 = 2$ involves 1, not prime itself.

Therefore, the statement is True.

Question 10

The function ( f(n) = \begin{cases} 1 & \text{if } n \text{ is odd} \ 2 & \text{if } n \text{ is even} \end{cases} ) is multiplicative.

A function $f$ is multiplicative if $f(ab) = f(a)f(b)$ for all pairs of coprime integers $a$ and $b$.

Let's check:

• If $a$ and $b$ are both odd, then $f(a) = 1$ and $f(b) = 1$, thus $f(ab) = 1$.
• If $a$ and $b$ are both even, then $f(a) = 2$ and $f(b) = 2$, thus $f(ab) = 2$ is not satisfied because the product of even numbers results in another even number and should return 2 as per the function definition.
• If one is odd and the other is even, $f(a)f(b) = 2$ and $f(ab) = 2$.

Since the condition does not hold consistently, the statement is False.

Would you like any further details or explanations on these problems? Here are five questions you might consider asking next:

1. How can the Euclidean algorithm be used to find the GCD?
2. What is the significance of Bézout's identity in number theory?
3. Can you provide an example of a multiplicative function?
4. How can we determine if a polynomial equation has integer solutions?
5. What are the properties of prime numbers in the context of quadratic equations?

Tip: Always verify your factorizations and results using different methods to ensure accuracy in your mathematical solutions.