Difference between revisions of "1990 AIME Problems/Problem 15"

Revision as of 11:03, 25 November 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:

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Find <math>a_{}^{}x^5 + b_{}y^5</math> if the real numbers <math>a_{}^{}</math>, <math>b_{}^{}</math>, <math>x_{}^{}</math>, and <math>y_{}^{}</math> satisfy the equations
+Find <math>a_{}^{}x^5 + b_{}y^5</math> if the [[real number]]s <math>a_{}^{}</math>, <math>b_{}^{}</math>, <math>x_{}^{}</math>, and <math>y_{}^{}</math> satisfy the [[equation]]s
 <cmath>ax + by = 3^{}_{},</cmath>
 <cmath>ax^2 + by^2 = 7^{}_{},</cmath>
@@ Line 30: / Line 30: @@
 == See also ==
 {{AIME box|year=1990|num-b=14|after=Last question}}
+[[Category:Intermediate Algebra Problems]]

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
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Difference between revisions of "1990 AIME Problems/Problem 15"

Revision as of 11:03, 25 November 2007

Problem

Solution

See also