How many values of r are there such that $\lfloor r \rfloor + r + \lfloor \frac{r}{2} \rfloor = 15.5?$
+506 note that \(\lfloor r \rfloor + \lfloor \frac{r}{2} \rfloor\) will be an integer, so r must be \(x+0.5\), where x is some positive integer.
Therefore,
\(\lfloor x+0.5 \rfloor + x + 0.5 + \lfloor \frac{x+0.5}{2} \rfloor = 15.5\\ x+ x + 0.5 + \lfloor \frac{x}{2} \rfloor = 15.5\\ 2x+\lfloor \frac{x}{2} \rfloor=15\)
If x is even, it reduces to \(\frac{5}{2}x=15\), which has one solution x=6, meaning that \(r=6\) is a solution.
If x is odd, we can write it as 2y+1, where y is some positive integer. The equation turns out to be:
\(2(2y+1)+y=15\\ 4y+2+y=15\\ 5y=13\)
since 13 is not a multiple of 5, no integer solutions exist for y, meaning that no odd solutions exist.
That means that 6 is the only solution, meaning that there is \(\boxed{1}\) value of r satisfying that condition.
