Let $ \alpha$ be the positive root of the equation $ x^{2} = 1991x + 1$. For natural numbers $ m$ and $ n$ define \[ m*n = mn + \lfloor\alpha m \rfloor \lfloor \alpha n\rfloor. \] Prove that for all natural numbers $ p$, $ q$, and $ r$, \[ (p*q)*r = p*(q*r). \]
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Tags: floor function, IMO Shortlist
hxy09
21.02.2009 15:43
Easily change it into the form $ [(\alpha - 1991)\{p\alpha\}\{q\alpha\}] = 0$ which is obviously true
a1k38
27.11.2009 16:50
I dont know why we have $ [(\alpha - 1991)\{p\alpha\}\{q\alpha\}] = 0$ ? :
MBGO
16.01.2013 22:24
in the fifth line you've been producted the term $[\alpha r]$ to the second statement.... am I right?