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# Floor function

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

May 11, 2022

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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