We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
71
2
avatar

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

 Aug 8, 2019
 #1
avatar+5798 
+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+23082 
+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

8 Online Users

avatar