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

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== Solution 2 ==
 
== Solution 2 ==
  
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Rearranging  the expression <math>N^2 + 7, we get  </math>N^2 + 7 = (N+4)(N-4) + 23.<math>
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Rearranging  the expression <math>N^2 + 7,</math> we get  <math>N^2 + 7 = (N+4)(N-4) + 23.</math>
  
−
Thus, </math> \frac{(N+4)(N-4) + 23}{N+4} = N - 4 + \frac{23}{N+4}.<math>
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Thus, <math> \frac{(N+4)(N-4) + 23}{N+4} = N - 4 + \frac{23}{N+4}.</math>
  
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Thus, </math>N+4 \equiv 0 \mod 23.<math>  
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Thus, <math>N+4 \equiv 0 \mod 23.</math>  
  
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Hence, </math>N= 23x -4,<math> for some arbitrary number </math>x.<math>
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Hence, <math>N= 23x -4,</math> for some arbitrary number <math>x.</math>
  
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Solving for </math>x<math> in  </math>23x-4<1990,<math> we find the answer is </math> x= 86,<math> or </math>\fbox{B}.$
+
Solving for <math>x</math> in  <math>23x-4<1990,</math> we find the answer is <math> x= 86,</math> or <math>\fbox{B}.</math>
  
 
~coolmath2017
 
~coolmath2017

Revision as of 21:29, 21 June 2020

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 (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
All AHSME Problems and Solutions

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