What is the Greatest Integer When Evaluating a Complex Fraction?

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The evaluation of the expression $$\left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor$$ results in a specific integer value. The calculations involve simplifying the complex fraction and applying the floor function to obtain the greatest integer. The consensus among participants confirms the accuracy of the approach taken by kaliprasad, emphasizing the importance of precise arithmetic operations in evaluating such expressions.

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Evaluate $$\left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor$$.
 
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anemone said:
Evaluate $$\left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor$$.

let x = 2015
so we get
$\left\lfloor\frac{(x-1)^3}{x(x+1)} + \frac{(x+1)^3}{(x-1)x}\right\rfloor$
$=\left\lfloor\frac{(x-1)^4+ (x+1)^4}{x(x+1)(x-1)}\right\rfloor$
$=\left\lfloor\frac{2(x^4+6x^2+ 2)}{x(x+1)(x-1)}\right\rfloor$
$=\left\lfloor\frac{2((x^2-1)(x^2+7)+9)}{x(x^2-1)}\right\rfloor$
$=\left\lfloor(\frac{2(x^2+7)}{x}+ \frac{18}{x(x^2-1)})\right\rfloor$
$=\left\lfloor(2x + \frac{14}{x} + \frac{18}{x(x^2-1)})\right\rfloor$
now as x = 2015 and so $\frac{14}{x} < \frac{1}{2}$ and $\frac{18}{x(x^2-1)} < \frac{1}{2}$ so
ans is 2x or 4030
 
Thanks for participating, kaliprasad! Just so you know that my approach is exactly the same as yours. (Smile)
 

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