For which integers $x$ is the quotient $\frac{x+11}{x+7}$ an integer? Find all solutions.

(L. Hozová)

Hint. Can you somehow simplify the given expression?
The given expression can be rewritten as follows ("whole part plus remainder"):

$$
\frac{x+11}{x+7}=\frac{(x+7)+4}{x+7}=\frac{x+7}{x+7}+\frac{4}{x+7}=1+\frac{4}{x+7} .
$$

Thus, $\frac{x+11}{x+7}$ is an integer if and only if $\frac{4}{x+7}$ is an integer, i.e., if and only if $x+7$ is a divisor of the number 4. The number 4 has six divisors: $-4,-2,-1,1,2$ and 4. The corresponding values of $x$ are 7 less. The number $\frac{x+11}{x+7}$ is an integer for $x$ equal to $-11,-9,-8,-6,-5$ or -3.

Another solution. To ensure the denominator is not zero, $x$ cannot be -7. By substituting integers near this value, we get:

| $x$ | -6 | -8 | -5 | -9 | -4 | -10 | -3 | -11 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\frac{x+11}{x+7}$ | 5 | -3 | 3 | -1 | $\frac{7}{3}$ | $-\frac{1}{3}$ | 2 | 0 | $\ldots$ |

As the distance of $x$ from -7 increases, we get non-integer values on the second row between 2 and 0, which approach 1 but never reach it: if $\frac{x+11}{x+7}=1$, then $x+11=x+7$, which is impossible. All valid options are thus contained in the given table. The number $\frac{x+11}{x+7}$ is an integer for $x$ equal to $-11,-9,-8,-6,-5$ or -3.

Note. Consider the previous argument as satisfactory, even though it is not perfect: to show correctly that for $x>-3$ or $x<-11$ the values of $\frac{x+11}{x+7}$ are between 2 and 0, probably exceeds the capabilities of solvers in this category.

This discussion relates to the monotonic behavior of the function $x \mapsto \frac{x+11}{x+7}$. The apparent confusion in the values of the function near $x=-7$ is related to the fact that the function is not defined at this point and its values are not constrained. Curious solvers may consider the behavior of this function, sketch its graph, and compare it with the previous conclusions.