Difference between revisions of "User:Btilm305/testproblem"

(test problem of the day)
 
(see what happens with a longer problem)
 
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Solve for <math>x</math>.  <math>x+12=15</math>.
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Solve for <math>x</math>.  <math>x+12=15</math>.  Bob and Janna blah blah blah go to the store.  Let me copy another problem... Let <math>x_1, x_2, ... , x_6</math> be non-negative real numbers such that <math>x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1</math>, and <math>x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\scriptstyle\frac{1}{540}}</math>. Let <math>p</math> and <math>q</math> be positive relatively prime integers such that <math>\frac{p}{q}</math> is the maximum possible value of
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<math>x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2</math>. Find <math>p+q</math>.

Latest revision as of 15:48, 3 June 2011

Solve for $x$. $x+12=15$. Bob and Janna blah blah blah go to the store. Let me copy another problem... Let $x_1, x_2, ... , x_6$ be non-negative real numbers such that $x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1$, and $x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\scriptstyle\frac{1}{540}}$. Let $p$ and $q$ be positive relatively prime integers such that $\frac{p}{q}$ is the maximum possible value of $x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2$. Find $p+q$.