+0  
 
0
55
1
avatar

How many values of r are there such that floor(4r) + r = 15.5?

 May 11, 2022
 #1
avatar+117498 
+2

I do not think there is any solution

 

How many values of r are there such that floor(4r) + r = 15.5?

 

Let  r = K+t       where  K is a positive integer   and   0<=t<1      (t for tiny)

 

\(\lfloor 4r\rfloor + r = 15.5\\ \lfloor 4(K+t)\rfloor + K+t = 15.5\\ \lfloor 4K+4t\rfloor + K+t = 15.5\\ \text{t is the only non integer so t=0.5}\\ \lfloor 4K+2\rfloor + K= 15\\ 4K+2+ K= 15\\ 5K= 13\\ K=2.6\\ \text{But K has to be an integer}\\ \text{That is a contradiction}\\ \text{Therefore there are no solution.}\)

 May 11, 2022

8 Online Users

avatar
avatar