Difference between revisions of "1990 AHSME Problems/Problem 19"
(Created page with "== Problem == For how many integers <math>N</math> between <math>1</math> and <math>1990</math> is the improper fraction <math>\frac{N^2+7}{N+4}</math> <math>\underline{not}</m...") |
(→Solution) |
||
(8 intermediate revisions by 5 users not shown) | |||
Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
− | <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. From that statement, we get that <math>m</math> divides <math>23</math> evenly. Since <math>23</math> is prime and we're looking for a GCD greater than 1, <math>m</math> must be <math>23</math>. Going back to our original statement, we can set <cmath>n+4=23x</cmath> for some positive integer x, 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>. |
== See also == | == See also == |
Latest revision as of 00:24, 9 August 2022
Problem
For how many integers between and is the improper fraction in lowest terms?
Solution
What we want to know is for how many is We start by setting for some arbitrary . This shows that evenly divides . Next we want to see under which conditions also divides . We know from the previous statement that and thus Next we simply add to get However, we also want which leads to from the previous statement. From that statement, we get that divides evenly. Since is prime and we're looking for a GCD greater than 1, must be . Going back to our original statement, we can set for some positive integer x, and Finally, we must find the largest such that This is a simple linear inequality for which the answer is , or .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.