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 "1986 AIME Problems/Problem 4"

m (→‎Solution)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Determine <math>\displaystyle 3x_4+2x_5</math> if <math>\displaystyle x_1</math>, <math>\displaystyle x_2</math>, <math>\displaystyle x_3</math>, <math>\displaystyle x_4</math>, and <math>\displaystyle x_5</math> satisfy the system of equations below.
+
Determine <math>3x_4+2x_5</math> if <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, <math>x_4</math>, and <math>x_5</math> satisfy the system of equations below.
−
<center><math>\displaystyle 2x_1+x_2+x_3+x_4+x_5=6</math></center>  
+
<center><math>2x_1+x_2+x_3+x_4+x_5=6</math></center>  
−
<center><math>\displaystyle x_1+2x_2+x_3+x_4+x_5=12</math></center>
+
<center><math>x_1+2x_2+x_3+x_4+x_5=12</math></center>
−
<center><math>\displaystyle x_1+x_2+2x_3+x_4+x_5=24</math></center>
+
<center><math>x_1+x_2+2x_3+x_4+x_5=24</math></center>
−
<center><math>\displaystyle x_1+x_2+x_3+2x_4+x_5=48</math></center>
+
<center><math>x_1+x_2+x_3+2x_4+x_5=48</math></center>
−
<center><math>\displaystyle x_1+x_2+x_3+x_4+2x_5=96</math></center>
+
<center><math>x_1+x_2+x_3+x_4+2x_5=96</math></center>
  
 
== Solution ==
 
== Solution ==
Line 17: Line 17:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 18:04, 4 July 2013

Problem

Determine $3x_4+2x_5$ if $x_1$, $x_2$, $x_3$, $x_4$, and $x_5$ satisfy the system of equations below.

$2x_1+x_2+x_3+x_4+x_5=6$
$x_1+2x_2+x_3+x_4+x_5=12$
$x_1+x_2+2x_3+x_4+x_5=24$
$x_1+x_2+x_3+2x_4+x_5=48$
$x_1+x_2+x_3+x_4+2x_5=96$

Solution

Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$. Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$, so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$.

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=1986_AIME_Problems/Problem_4&oldid=54885"