Find the all arranged triplets ,from positive real numbers ( a,b,c)
that: (a)bc=3,a(b)c=4,ab(c)=5,which (x) is the greatest integer less than or equal to x
Label the original equations (1), (2), and (3) in that order.
First, note that $\displaystyle a,b,c \ge 1$ (otherwise $\displaystyle \lfloor a \rfloor = 0$, etc., contradiction).
Multiplying all three equations, we get $\displaystyle a^2b^2c^2 = \frac{60}{\lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor} \Rightarrow abc = \frac{\sqrt{60}}{\sqrt{\lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor}}$.
We can prove that $\displaystyle a,b,c < 3$. Assume that $\displaystyle \lfloor b \rfloor = 3$. In order to satisfy (1), b must equal 3, otherwise c would have to be less than 1. If b = 3, $\displaystyle \lfloor a \rfloor = c = 1$. Solving for a via (2) and (3), we obtain $\displaystyle a = \frac{4}{3}$ and $\displaystyle a = \frac{5}{3}$, no solution. Similarly, if $\displaystyle \lfloor c \rfloor = 3$, we get the same contradiction. If $\displaystyle \lfloor a \rfloor = 3$, $\displaystyle b = c = 1$, no solutions.
Therefore, $\displaystyle \lfloor a \rfloor, \lfloor b \rfloor, \lfloor c \rfloor \in {1,2}$. It follows that $\displaystyle \lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor \in {1,2,4}$. I'll let you do the casework (I obtained at least two solutions btw).
(There should be {} with 1,2 and 1,2,4, but LaTeX interprets the brackets differently, and idk how to input them otherwise).