19. The expression $\frac{7 n+12}{2 n+3}$ takes integer values for certain integer values of $n$.
What is the sum of all such integer values of the expression?
A 4
B 8
C 10
D 12
E 14
Solution
E

We have
$$
\frac{7 n+12}{2 n+3}=\frac{7}{2}+\frac{3}{2(2 n+3)}=\frac{1}{2}\left(7+\frac{3}{2 n+3}\right) .
$$

It follows that $\frac{7 n+12}{2 n+3}$ is an integer provided that $7+\frac{3}{2 n+3}$ is an even integer and hence provided that $\frac{3}{2 n+3}$ is an odd integer.
Now $\frac{3}{2 n+3}$ is an integer provided that $2 n+3$ is a factor of 3 .
Therefore the only possible values of $2 n+3$ are $-3,-1,1$ and 3 . In all these cases $\frac{3}{2 n+3}$ is a factor of 3 and hence is odd.
When $2 n+3=-3$, we have $n=-3$ and $\frac{7 n+12}{2 n+3}=\frac{1}{2}\left(7+\frac{3}{2 n+3}\right)=\frac{1}{2}\left(7+\frac{3}{-3}\right)=\frac{1}{2}(7-1)=3$.
When $2 n+3=-1$, we have $n=-2$ and $\frac{7 n+12}{2 n+3}=\frac{1}{2}\left(7+\frac{3}{2 n+3}\right)=\frac{1}{2}\left(7+\frac{3}{-1}\right)=\frac{1}{2}(7-3)=2$.
When $2 n+3=1$, we have $n=-1$ and $\frac{7 n+12}{2 n+3}=\frac{1}{2}\left(7+\frac{3}{2 n+3}\right)=\frac{1}{2}\left(7+\frac{3}{1}\right)=\frac{1}{2}(7+3)=5$.
When $2 n+3=3$, we have $n=0$ and $\frac{7 n+12}{2 n+3}=\frac{1}{2}\left(7+\frac{3}{2 n+3}\right)=\frac{1}{2}\left(7+\frac{3}{3}\right)=\frac{1}{2}(7+1)=4$.
We see that $n$ is an integer whenever $2 n+3$ is equal to either $-3,-1,1$ or 3 . Therefore the sum of the integer values of $\frac{7 n+12}{2 n+3}$ that correspond to integer values of $n$ is $3+2+5+4=14$.