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Compute \(\frac{\lfloor \sqrt[4]{1} \rfloor \cdot \lfloor \sqrt[4]{3} \rfloor \cdot \lfloor \sqrt[4]{5} \rfloor \dotsm \lfloor \sqrt[4]{2015} \rfloor}{\lfloor \sqrt[4]{2} \rfloor \cdot \lfloor \sqrt[4]{4} \rfloor \cdot \lfloor \sqrt[4]{6} \rfloor \dotsm \lfloor \sqrt[4]{2016} \rfloor}.\)

 Dec 2, 2018
edited by Guest  Dec 2, 2018
 #1
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deleted. Sorry, misread the "floor function"

 Dec 2, 2018
edited by Guest  Dec 3, 2018
 #2
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deleted.

 Dec 2, 2018
edited by Guest  Dec 2, 2018
edited by Guest  Dec 3, 2018
 #3
avatar+98172 
+1

Note that 

2^4 = 16

3^4 = 81

4^4 = 256

5^4 = 625

6^4 = 1296

7^4 = 2401

 

So each floor from (1)^.25 to ( 15)^.25  = 1

And floor (16)^.25 = 2

 

And each  floor  from (17)^.25 to (80)^.25 = 2

And floor (81)^.25 = 3

 

And each floor  from (82)^.25 to (255)^.25 = 3

And floor (256)^.25 = 4

 

And each floor  from (257)^.25 to (624)^.25 = 4

And floor (625)^.25 = 5

 

And each floor  from (626)^.25 to (1295)^.25 = 5

And floor (1296)^.25 = 6

 

And each floor  from (1297)^.25 to (2016)^.25 =  6

 

 

So...note the pattern.....

 

(1)^8 *  (2)^32 * (3)^88 * (4)^184 * ....               1 * 3 * ......

______________________________        =  ___________

(1)^7 * (2)^33 * (3)^87* (4)^185 * .......               2 * 4 * .....

 

 

1 *  3 *  5 * 

__________  =

  2 * 4 * 6

 

 

15           

____ =     

 48   

 

 

5

__

16

 

     

 

 

cool cool cool

 Dec 2, 2018
edited by CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
 #4
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How can the top be greater than the bottom though

Guest Dec 3, 2018
 #5
avatar+98172 
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Sorry.....I shouldn't have included "7' in the numerator...we only have a floor of 6 on (2016)^.25

 

I believe tthe final answer is correct....thanks for noticing my error...

 

 

cool cool cool

CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
 #6
avatar+18 
+1

I think your calculation is incorrect, the answer I got is \(\frac{1}{\lfloor \sqrt[4]{2016} \rfloor}\)

HelloWorld  Dec 3, 2018
edited by HelloWorld  Dec 3, 2018

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