+0  
 
0
152
2
avatar

Compute \( \left\lfloor \frac{2007! + 2004!}{2006! + 2005!}\right\rfloor. \)

 Aug 8, 2019
 #1
avatar+6046 
+1

\(\left \lfloor \dfrac{2007! + 2004!}{2006!+2005!}\right\rfloor = \\ \left \lfloor \dfrac{2004!(1+2005\cdot 2006 \cdot 2007)}{2004!\cdot 2005(1+2006)}\right\rfloor = \\ \left \lfloor \dfrac{1+2005\cdot 2006 \cdot 2007}{2005(1+2006)}\right\rfloor = \\ \left \lfloor \dfrac{\frac{1}{2005}+2006 \cdot 2007}{1+2006}\right\rfloor = \\ \left \lfloor \dfrac{\frac{1}{2005\cdot 2006}+2007}{\frac{1}{2006}+1}\right\rfloor = 2006\\\)

 

The last equality comes from the first term in the numerator being so small we can ignore it, 

and the denominator being slightly greater than 1 so that the floor is 2006 rather than 2007.

.
 Aug 9, 2019
edited by Rom  Aug 9, 2019
 #2
avatar+23850 
+2

Compute  \(\Bigg\lfloor \dfrac{2007! + 2004!}{2006! + 2005!}\Bigg\rfloor\)

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ \Bigg\lfloor \dfrac{2007! + 2004!}{2006! + 2005!}\Bigg \rfloor }\\ &=& \Bigg\lfloor \dfrac{2007!}{2006! + 2005!} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006!~2007}{2005!~(1+2006)} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006!~2007}{2005!~2007} + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2006! }{2005! } + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor \dfrac{2005!~2006 }{2005! } + \dfrac{2004!}{2006! + 2005!} \Bigg \rfloor \\ &=& \Bigg\lfloor 2006 + \underbrace{\dfrac{2004!}{2006! + 2005!}}_{<1} \Bigg \rfloor \\ &=& \mathbf{2006} \\ \hline \end{array}\)

 

laugh

 Aug 9, 2019

38 Online Users