Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1969 AHSME Problems/Problem 19"

(Created page with "== Problem == The number of distinct ordered pairs <math>(x,y)</math> where <math>x</math> and <math>y</math> have positive integral values satisfying the equation <math>x^4y^4-...")
 
(→‎Solution)
 
(4 intermediate revisions by 2 users not shown)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{D}</math>
+
Let <math>(xy)^2=a</math>. The expression given is equal to <math>a^2-10a+9=0</math>, which can be factored as <math>(a-9)(a-1)=0</math>. Thus, we have <math>a=(xy)^2=9</math> and <math>a=(xy)^2=1</math>. Because <math>x</math> and <math>y</math> are positive, we can eliminate the possibilities where <math>xy</math> is negative. From here it is easy to see that the only integral pairs of <math>x</math> and <math>y</math> are <math>(3, 1), (1, 3)</math>, and <math>(1, 1)</math>. The answer is <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME 35p box|year=1969|num-b=18|num-a=20}}   
 
{{AHSME 35p box|year=1969|num-b=18|num-a=20}}   
  
−
[[Category: Intermediate Algebra Problems]]
+
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:14, 10 July 2015

Problem

The number of distinct ordered pairs $(x,y)$ where $x$ and $y$ have positive integral values satisfying the equation $x^4y^4-10x^2y^2+9=0$ is:

$\text{(A) } 0\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 12\quad \text{(E) } \infty$

Solution

Let $(xy)^2=a$. The expression given is equal to $a^2-10a+9=0$, which can be factored as $(a-9)(a-1)=0$. Thus, we have $a=(xy)^2=9$ and $a=(xy)^2=1$. Because $x$ and $y$ are positive, we can eliminate the possibilities where $xy$ is negative. From here it is easy to see that the only integral pairs of $x$ and $y$ are $(3, 1), (1, 3)$, and $(1, 1)$. The answer is $\fbox{B}$.

See also

1969 AHSC (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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1969_AHSME_Problems/Problem_19&oldid=71078"