16.8

Algebra Level 4

n = 0 2016 n e n π = ? \large \sum_{n=0}^{2016} \left \lfloor \dfrac{n-e}{n-\pi} \right \rfloor = \, ?


The answer is 2011.

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1 solution

n = 0 2016 n e n π = n = 0 2016 1 + π e n π = 2017 + n = 0 2016 π e n π π e n π = { 1 x = 0 , 1 , 2 3 x = 3 0 x > 3 2017 + n = 0 2016 π e n π = 2017 1 1 1 3 = 2011 \sum_{n=0}^{2016} \left \lfloor \dfrac{n-e}{n-\pi} \right \rfloor= \sum_{n=0}^{2016} \left \lfloor 1+\frac{\pi-e}{n-\pi} \right \rfloor=2017+ \sum_{n=0}^{2016} \left \lfloor \dfrac{\pi-e}{n-\pi} \right \rfloor\\\\ \left \lfloor \dfrac{\pi-e}{n-\pi}\right\rfloor=\begin{cases}-1 & x = 0,1,2\\-3 & x = 3\\0&x>3\end{cases}\\\therefore 2017+ \sum_{n=0}^{2016} \left \lfloor \dfrac{\pi-e}{n-\pi} \right \rfloor=2017-1-1-1-3=2011

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