Professor M. A. Luco wrote on the board the expression:

$$
\frac{n^{2}-5 n+4}{n-4}
$$

Then, he tells the students that $n$ can be any natural number, except for 4.

a) What is the value of the expression for $n=1$?

b) Marcos substituted $n$ with a natural number and found that the value of the expression is 5. Which number did Marcos substitute for $n$?

c) Which natural numbers cannot be the numerical value of the expression?
Solution

a) For $n=1$, we have:

$$
\begin{aligned}
\frac{n^{2}-5 n+4}{n-4} & =\frac{1^{2}-5 \cdot 1+4}{1-4} \\
& =\frac{1-5+4}{-3} \\
& =0
\end{aligned}
$$

b) Setting the expression equal to 5, we get:

$$
\begin{aligned}
\frac{n^{2}-5 n+4}{n-4} & =5 \\
n^{2}-5 n+4 & =5(n-4) \\
n^{2}-10 n+24 & =0 \\
n & =\frac{10 \pm \sqrt{(-10)^{2}-4 \cdot 1 \cdot 24}}{2 \cdot 1} \\
& =\frac{10 \pm \sqrt{100-96}}{2} \\
& =\frac{10 \pm 2}{2} \\
& =5 \pm 1
\end{aligned}
$$

Therefore, Marcos substituted $n$ with 6, since $n \neq 4$.

c) Factoring the numerator and remembering that $n \neq 4$, we have:

$$
\frac{n^{2}-5 n+4}{n-4}=\frac{(n-4)(n-1)}{n-4}=n-1
$$

Since $n \neq 4$, then $n-1 \neq 3$. Therefore, the expression can take any natural number except 3.