# Finding the triplets

• Jun 21st 2012, 01:56 PM
Mhmh96
Finding the triplets
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
• Jun 21st 2012, 04:39 PM
richard1234
Re: Finding the triplets
Label the original equations (1), (2), and (3) in that order.

First, note that $a,b,c \ge 1$ (otherwise $\lfloor a \rfloor = 0$, etc., contradiction).

Multiplying all three equations, we get $a^2b^2c^2 = \frac{60}{\lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor} \Rightarrow abc = \frac{\sqrt{60}}{\sqrt{\lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor}}$.

We can prove that $a,b,c < 3$. Assume that $\lfloor b \rfloor = 3$. In order to satisfy (1), b must equal 3, otherwise c would have to be less than 1. If b = 3, $\lfloor a \rfloor = c = 1$. Solving for a via (2) and (3), we obtain $a = \frac{4}{3}$ and $a = \frac{5}{3}$, no solution. Similarly, if $\lfloor c \rfloor = 3$, we get the same contradiction. If $\lfloor a \rfloor = 3$, $b = c = 1$, no solutions.

Therefore, $\lfloor a \rfloor, \lfloor b \rfloor, \lfloor c \rfloor \in {1,2}$. It follows that $\lfloor a \rfloor \lfloor b \rfloor \lfloor c \rfloor \in {1,2,4}$. 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).
• Jun 21st 2012, 04:52 PM
Reckoner
Re: Finding the triplets
Quote:

Originally Posted by richard1234
(There should be {} with 1,2 and 1,2,4, but LaTeX interprets the brackets differently, and idk how to input them otherwise).

Nice solution. You can get braces with \{ and \}: $\lfloor a\rfloor\lfloor b\rfloor\lfloor c\rfloor\in\{1,2,4\}$
• Jun 22nd 2012, 12:22 AM
richard1234
Re: Finding the triplets
Quote:

Originally Posted by Reckoner
Nice solution. You can get braces with \{ and \}: $\lfloor a\rfloor\lfloor b\rfloor\lfloor c\rfloor\in\{1,2,4\}$

Ah, thanks.