+118608 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)
\(\lfloor 4r\rfloor + r = 15.5\\ \lfloor 4(K+t)\rfloor + K+t = 15.5\\ \lfloor 4K+4t\rfloor + K+t = 15.5\\ \text{t is the only non integer so t=0.5}\\ \lfloor 4K+2\rfloor + K= 15\\ 4K+2+ K= 15\\ 5K= 13\\ K=2.6\\ \text{But K has to be an integer}\\ \text{That is a contradiction}\\ \text{Therefore there are no solution.}\)
