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Difference between revisions of "2023 WSMO Tiebreaker Round Problems/Problem 2"

(Created page with "==Problem== If <math>x^3+y^3=76895, x<y</math> and <math>x+y=65,</math> find <math>xy.</math> ==Solution==")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
Note that
 +
<cmath>\begin{align*}
 +
xy &= \frac{xy(x+y)}{(x+y)}\\
 +
&= \frac{1}{3}\left(\frac{3x^2y+3xy^2}{x+y}\right)\\
 +
&= \frac{1}{3}\left(\frac{(x^3+3x^2y+3xy^2+y^3)-(x^3+y^3)}{x+y}\right)\\
 +
&= \frac{1}{3}\left(\frac{(x+y)^3-(x^3+y^3)}{x+y}\right)\\
 +
&= \frac{1}{3}\left(\frac{65^3-76895}{65}\right)\\
 +
&= \frac{65^2-1183}{3} = \boxed{1014}.
 +
\end{align*}</cmath>
 +
 +
~pinkpig

Latest revision as of 10:46, 15 September 2025

Problem

If $x^3+y^3=76895, x<y$ and $x+y=65,$ find $xy.$

Solution

Note that \begin{align*} xy &= \frac{xy(x+y)}{(x+y)}\\ &= \frac{1}{3}\left(\frac{3x^2y+3xy^2}{x+y}\right)\\ &= \frac{1}{3}\left(\frac{(x^3+3x^2y+3xy^2+y^3)-(x^3+y^3)}{x+y}\right)\\ &= \frac{1}{3}\left(\frac{(x+y)^3-(x^3+y^3)}{x+y}\right)\\ &= \frac{1}{3}\left(\frac{65^3-76895}{65}\right)\\ &= \frac{65^2-1183}{3} = \boxed{1014}. \end{align*}

~pinkpig

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