Consider the fraction

$$
t=\frac{5 n+9}{n-3}
$$

if $n$ is an integer greater than 3, and determine:

$1^{0}.$ for which values of $n$ will $t$ be an integer,

$2^{0}.$ for which values of $n$ will the fraction be an irreducible fraction.
$1^{0}$. $t$ is an integer if the numerator is a multiple of the denominator (an integer). After performing the division,

$$
t=\frac{5(n-3)+24}{n-3}=5+\frac{24}{n-3}
$$

It is necessary that $n-3$ is a divisor of 24, that is,

$$
n-3=1,2,3,4,6,8,12,24
$$

and thus

$$
\begin{aligned}
& n=\quad 4, \quad 5, \quad 6, \quad 7, \quad 9, \quad 11, \quad 15, \quad 27 \\
& t=29, \quad 17, \quad 13, \quad 11, \quad 9, \quad 8, \quad 7, \quad 6
\end{aligned}
$$

$2^{0}$. For $\frac{24}{n-3}$ to be irreducible, it is necessary that the denominator does not contain the prime factors of the numerator, 2 and 3. The denominator is not divisible by 2 if $n$ is an even number.

If $n$ is divisible by 3, then the numerator and the denominator have a common divisor, and the fraction can be simplified; however, if $n$ is an even number that is not a multiple of 3, then the fraction is irreducible. (We exclude the case $n=4$; this belongs to the first part $1^{0}$.)

Therefore, $n$ must be a number of the form $n=6k+2$, where $k$ is any ordinary integer, but $n>6$.