How many values of r are there such that $\lfloor r/4 \rfloor + r = 15.5?$
+9519 Moving r to the other side gives \(\left\lfloor\dfrac r4\right\rfloor = 15.5 -r \).
Then 15.5 - r is an integer. Then r = k + 0.5 for some integer k.
Then \(\left\lfloor\dfrac {k + \frac12}4\right\rfloor = 15 - k\).
That means \(15-k\leq \dfrac{k + \frac12}4 < 16 - k\) by definition of floor function. Now it becomes a compound inequality. To proceed, solve the compound inequality and list its integer solutions.
