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

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== Solution == | == Solution == | ||

− | { | + | <math>(ax^2+by^2)(x+y)=7(x+y)=ax^3+by^3+(ax+by)xy=16+3xy</math> |

+ | |||

+ | <math>(ax^3+by^3)(x+y)=16(x+y)=ax^4+by^4+(ax^2+by^2)xy=42+7xy</math> | ||

+ | |||

+ | x+y=a | ||

+ | |||

+ | xy=b | ||

+ | |||

+ | <math>3b=7a-16</math> | ||

+ | |||

+ | <math>7b=16a-42</math> | ||

+ | |||

+ | <math>21b=49a-112</math> | ||

+ | |||

+ | <math>21b=48a-126</math> | ||

+ | |||

+ | <math>49a-112=48a-126</math> | ||

+ | |||

+ | <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}} |