1990 AHSME Problems/Problem 19

Revision as of 22:29, 21 June 2020 by Coolmath2017 (talk | contribs) (Solution 2)

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

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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