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 "1990 AIME Problems/Problem 15"
(→‎Solution: solution by towersfreak2006)
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
−
<math>(ax^2+by^2)(x+y)=7(x+y)=ax^3+by^3+(ax+by)xy=16+3xy</math>
+
Set <math>S = (x + y)</math> and <math>P = xy</math>. Then the relationship
  
−
<math>(ax^3+by^3)(x+y)=16(x+y)=ax^4+by^4+(ax^2+by^2)xy=42+7xy</math>
+
<cmath>(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})</cmath>
  
−
x+y=a
+
can be exploited:
  
−
xy=b
+
<cmath>\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\
 +
(ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}</cmath>
  
−
<math>3b=7a-16</math>
+
Therefore:
  
−
<math>7b=16a-42</math>
+
<cmath>\begin{eqnarray*}7S & = & 16 + 3P \\
 +
16S & = & 42 + 7P\end{eqnarray*}</cmath>
  
−
<math>21b=49a-112</math>
+
Consequently, <math>S = - 14</math> and <math>P = - 38</math>. Finally:
  
−
<math>21b=48a-126</math>
+
<cmath>\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\
−
 
+
(42)(S) & = & (ax^5 + by^5) + (P)(16) \\
−
<math>49a-112=48a-126</math>
+
(42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\
−
 
+
ax^5 + by^5 & = & \boxed{20}\end{eqnarray*}</cmath>
−
<math>a=-126+112=-14</math>
 
−
 
 
−
<math>3b=-98-16=-114 \Rightarrow b=-38</math>
 
−
 
 
−
<math>(ax^4+by^4)(x+y)=42a=ax^5+by^5+(ax^3+by^3)xy=ax^5+by^5+16b</math>
 
−
 
 
−
<math>42a-16b=ax^5+by^5</math>
 
−
 
 
−
<math>-14*42+38*16=\boxed{050}</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=14|after=Last question}}
 
{{AIME box|year=1990|num-b=14|after=Last question}}

Revision as of 20:42, 26 October 2007

Problem

Find $a_{}^{}x^5 + b_{}y^5$ if the real numbers $a_{}^{}$, $b_{}^{}$, $x_{}^{}$, and $y_{}^{}$ satisfy the equations \[ax + by = 3^{}_{},\] \[ax^2 + by^2 = 7^{}_{},\] \[ax^3 + by^3 = 16^{}_{},\] \[ax^4 + by^4 = 42^{}_{}.\]

Solution

Set $S = (x + y)$ and $P = xy$. Then the relationship

\[(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})\]

can be exploited:

\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}

Therefore:

\begin{eqnarray*}7S & = & 16 + 3P \\ 16S & = & 42 + 7P\end{eqnarray*}

Consequently, $S = - 14$ and $P = - 38$. Finally:

\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ ax^5 + by^5 & = & \boxed{20}\end{eqnarray*}

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&oldid=18909"