Problem 180

What we want to know is for how many $n$ is $$\gcd(n^2+7, n+4) > 1.$$ We start by setting $$n+4 \equiv 0 \mod m$$ for some arbitrary $m$. This shows that $m$ evenly divides $n+4$. Next we want to see under which conditions $m$ also divides $n^2 + 7$. We know from the previous statement that $$n \equiv -4 \mod m$$ and thus $$n^2 \equiv (-4)^2 \equiv 16 \mod m.$$ Next we simply add $7$ to get $$n^2 + 7 \equiv 23 \mod m.$$ However, we also want $$n^2 + 7 \equiv 0 \mod m$$ which leads to $$n^2 + 7\equiv 23 \equiv 0 \mod m$$ from the previous statement. From that statement, we get that $m$ divides $23$ evenly. Since $23$ is prime and we're looking for a GCD greater than 1, $m$ must be $23$. Going back to our original statement, we can set $$n+4=23x$$ for some positive integer x, and $$n=23x-4.$$ Finally, we must find the largest $x$ such that $$23x-4<1990.$$ This is a simple linear inequality for which the answer is $x=86$, or $\fbox{B}$.