Difference between revisions of "Mock AIME II 2012 Problems/Problem 11"

Revision as of 02:17, 5 April 2012

Problem

There exist real values of $a$ and $b$ such that $a+b=n$ , $a^2+b^2=2n$ , and $a^3+b^3=3n$ for some value of $n$ . Let $S$ be the sum of all possible values of $a^4+b^4$ . Find $S$ .

Solution

First, if $n=0$ , then $a^2+b^2=0\implies a=b=0$ . We now assume that $n\ne 0$ . Now, note that $2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab$ . Also, we have $(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies 2n+2(2n-3)=n^2\implies n^2-6n+6=0$ .

Next, $3n^2=(a+b)(a^3+b^3)=a^4+b^4+ab(a^2+b^2)=a^4+b^4+(2n-3)(2n)\implies a^4+b^4=-n^2+6n$ . But we know $n^2-6n+6=0$ , so $-n^2+6n=6$ .

Since the only possible values of $a^4+b^4$ are $0$ and $6$ , our final answer is $\boxed{006}$ .

(It is easy to check that there exists $a, b, n$ satisfying the equations.)

@@ Line 4: / Line 4: @@
 ==Solution==
- First, if <math>n=0</math>, then <math>a^2+b^2=0\implies a=b=0</math>. We now assume that <math>n\ne 0</math>.
+First, if <math>n=0</math>, then <math>a^2+b^2=0\implies a=b=0</math>. We now assume that <math>n\ne 0</math>.
 Now, note that <math>2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab</math>.
 Also, we have <math>(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies 2n+2(2n-3)=n^2\implies n^2-6n+6=0</math>.

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Difference between revisions of "Mock AIME II 2012 Problems/Problem 11"

Revision as of 02:17, 5 April 2012

Problem

Solution