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

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1990 AHSME Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

See also