+0

# help with floor

0
50
1

How many values of r are there such that $\lfloor r \rfloor + r + \lfloor \frac{r}{2} \rfloor = 15.5?$

Aug 6, 2021

#1
+436
+1

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.

Aug 6, 2021