How many values of r are there such that $\lfloor r \rfloor + r = 15.5 + \lfloor 4r \rfloor$ ?

Guest Jun 15, 2022

#1**+1 **

That means \(2\lfloor r \rfloor + 2r = 31 + 2\lfloor 4r \rfloor\).

Now, the right-hand side of the equation is an integer and \(2\lfloor r \rfloor\) is an integer. That means \(2r\) is also an integer.

Case 1: Let \(r = x + 0.5\), where x is an integer.

Then

\(2x + \dfrac12 = 15.5 + 4x + 2\\ 2x = 17 + 4x\\ 2x = -17\\ x = -\dfrac{17}2\)

which is not an integer. Therefore, there are no roots of such form.

Case 2: \(r\) is an integer.

\(2r = 15.5 + 4r\)

\(r = -\dfrac{31}4\)

which is not an integer.

Therefore, there are no such \(r\) satisfying the equation.

MaxWong Jun 15, 2022