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

Revision as of 08:51, 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

$3b=7a-16$

$7b=16a-42$

$21b=49a-112$

$21b=48a-126$

$49a-112=48a-126$

$a=-126+112=-14$

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

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

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

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

@@ 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>
+<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 ==
 {{AIME box|year=1990|num-b=14|after=Last question}}

1990 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1990 AIME Problems/Problem 15"

Revision as of 08:51, 16 October 2007

Problem

Solution

See also