2023 WSMO Tiebreaker Round Problems/Problem 2

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

Page

Toolbox

Search

2023 WSMO Tiebreaker Round Problems/Problem 2

Problem

Solution