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

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== Solution ==
 
== Solution ==
{{solution}}
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<math>(ax^2+by^2)(x+y)=7(x+y)=ax^3+by^3+(ax+by)xy=16+3xy</math>
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<math>(ax^3+by^3)(x+y)=16(x+y)=ax^4+by^4+(ax^2+by^2)xy=42+7xy</math>
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x+y=a
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xy=b
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<math>3b=7a-16</math>
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<math>7b=16a-42</math>
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<math>21b=49a-112</math>
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<math>21b=48a-126</math>
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<math>49a-112=48a-126</math>
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<math>a=-126+112=-14</math>
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<math>3b=-98-16=-114 \Rightarrow b=-38</math>
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<math>(ax^4+by^4)(x+y)=42a=ax^5+by^5+(ax^3+by^3)xy=ax^5+by^5+16b</math>
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<math>42a-16b=ax^5+by^5</math>
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<math>-14*42+38*16=\boxed{050}</math>
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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}}

Revision as of 09:51, 16 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

$(ax^2+by^2)(x+y)=7(x+y)=ax^3+by^3+(ax+by)xy=16+3xy$

$(ax^3+by^3)(x+y)=16(x+y)=ax^4+by^4+(ax^2+by^2)xy=42+7xy$

x+y=a

xy=b

$3b=7a-16$

$7b=16a-42$

$21b=49a-112$

$21b=48a-126$

$49a-112=48a-126$

$a=-126+112=-14$

$3b=-98-16=-114 \Rightarrow b=-38$

$(ax^4+by^4)(x+y)=42a=ax^5+by^5+(ax^3+by^3)xy=ax^5+by^5+16b$

$42a-16b=ax^5+by^5$

$-14*42+38*16=\boxed{050}$

See also

1990 AIME (ProblemsAnswer KeyResources)
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