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

(Solution: solution by towersfreak2006)
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== Problem ==
 
== 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
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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 + by = 3^{}_{},</cmath>
 
<cmath>ax^2 + by^2 = 7^{}_{},</cmath>
 
<cmath>ax^2 + by^2 = 7^{}_{},</cmath>
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== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=14|after=Last question}}
 
{{AIME box|year=1990|num-b=14|after=Last question}}
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[[Category:Intermediate Algebra Problems]]

Revision as of 12:03, 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*}

See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
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