# Difference between revisions of "1990 AHSME Problems/Problem 19"

## Problem

For how many integers $N$ between $1$ and $1990$ is the improper fraction $\frac{N^2+7}{N+4}$ $\underline{not}$ in lowest terms? $\text{(A) } 0\quad \text{(B) } 86\quad \text{(C) } 90\quad \text{(D) } 104\quad \text{(E) } 105$

## Solution 1

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. Since from that statement $23$ divides $m$ evenly, $m$ must be of the form $23x$, for some arbitrary integer $x$. After this, we can set $$n+4=23x$$ 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 $\fbox{B}$.

## Solution 2

Rearranging the expression $N^2 + 7,$ we get $N^2 + 7 = (N+4)(N-4) + 23.$

Thus, $\frac{(N+4)(N-4) + 23}{N+4} = N - 4 + \frac{23}{N+4}.$

Thus, $N+4 \equiv 0 \mod 23.$

Hence, $N= 23x -4,$ for some arbitrary number $x.$

Solving for $x$ in $23x-4<1990,$ we find the answer is $x= 86,$ or $\fbox{B}.$

~coolmath2017

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