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

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\text{(E) } 105</math>
 
\text{(E) } 105</math>
  
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== Solution ==
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== Solution 1 ==
 
What we want to know is for how many <math>n</math> is <cmath>\gcd(n^2+7, n+4) > 1.</cmath> We start by setting <cmath>n+4 \equiv 0 \mod m</cmath> for some arbitrary <math>m</math>. This shows that <math>m</math> evenly divides <math>n+4</math>. Next we want to see under which conditions <math>m</math> also divides <math>n^2 + 7</math>.  We know from the previous statement that <cmath>n \equiv -4 \mod m</cmath> and thus <cmath>n^2 \equiv (-4)^2 \equiv 16 \mod m.</cmath> Next we simply add <math>7</math> to get <cmath>n^2 + 7 \equiv 23 \mod m.</cmath> However, we also want <cmath>n^2 + 7 \equiv 0 \mod m</cmath> which leads to <cmath>n^2 + 7\equiv 23 \equiv 0 \mod m</cmath> from the previous statement. Since from that statement <math>23</math> divides <math>m</math> evenly, <math>m</math> must be of the form <math>23x</math>, for some arbitrary integer <math>x</math>. After this, we can set <cmath>n+4=23x</cmath> and <cmath>n=23x-4.</cmath> Finally, we must find the largest <math>x</math> such that <cmath>23x-4<1990.</cmath> This is a simple linear inequality for which the answer is <math>x=86</math>, or <math>\fbox{B}</math>.
 
What we want to know is for how many <math>n</math> is <cmath>\gcd(n^2+7, n+4) > 1.</cmath> We start by setting <cmath>n+4 \equiv 0 \mod m</cmath> for some arbitrary <math>m</math>. This shows that <math>m</math> evenly divides <math>n+4</math>. Next we want to see under which conditions <math>m</math> also divides <math>n^2 + 7</math>.  We know from the previous statement that <cmath>n \equiv -4 \mod m</cmath> and thus <cmath>n^2 \equiv (-4)^2 \equiv 16 \mod m.</cmath> Next we simply add <math>7</math> to get <cmath>n^2 + 7 \equiv 23 \mod m.</cmath> However, we also want <cmath>n^2 + 7 \equiv 0 \mod m</cmath> which leads to <cmath>n^2 + 7\equiv 23 \equiv 0 \mod m</cmath> from the previous statement. Since from that statement <math>23</math> divides <math>m</math> evenly, <math>m</math> must be of the form <math>23x</math>, for some arbitrary integer <math>x</math>. After this, we can set <cmath>n+4=23x</cmath> and <cmath>n=23x-4.</cmath> Finally, we must find the largest <math>x</math> such that <cmath>23x-4<1990.</cmath> This is a simple linear inequality for which the answer is <math>x=86</math>, or <math>\fbox{B}</math>.
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== 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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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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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}.$
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~coolmath2017
  
 
== See also ==
 
== See also ==

Revision as of 21:27, 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
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