Floor Question System of Equations

We are tasked with finding all possible values of \( b \) that satisfy the following system of equations involving the floor functions:

 

\[
\lfloor a \rfloor \cdot b \cdot c = 3 \tag{1}
\]

\[
a \cdot \lfloor b \rfloor \cdot c = 4 \tag{2}
\]

\[
a \cdot b \cdot \lfloor c \rfloor = 5 \tag{3}
\]

 

where \( a \), \( b \), and \( c \) are positive real numbers. We will solve this step by step by exploring possible values for \( a \), \( b \), and \( c \).

 

### Solution By Steps

 

#### Step 1: Analyze the first equation

From Equation (1):

 

\[
\lfloor a \rfloor \cdot b \cdot c = 3
\]

 

The floor function \( \lfloor a \rfloor \) represents the greatest integer less than or equal to \( a \). Since \( a \) is positive, the possible values of \( \lfloor a \rfloor \) are integers.

 

We first assume \( \lfloor a \rfloor = 1 \), which is the smallest possible value because \( a > 0 \). Substituting this into Equation (1):

 

\[
1 \cdot b \cdot c = 3
\]

 

Thus, we have:

 

\[
b \cdot c = 3 \tag{4}
\]

 

#### Step 2: Analyze the second equation

Now, consider Equation (2):

 

\[
a \cdot \lfloor b \rfloor \cdot c = 4
\]

 

We already know \( b \cdot c = 3 \) from Equation (4). To satisfy this equation, we explore possible values of \( \lfloor b \rfloor \).

 

Let’s assume \( \lfloor b \rfloor = 1 \) first. Substituting into Equation (2):

 

\[
a \cdot 1 \cdot c = 4
\]

 

So,

 

\[
a \cdot c = 4 \tag{5}
\]

 

#### Step 3: Analyze the third equation

Now consider Equation (3):

 

\[
a \cdot b \cdot \lfloor c \rfloor = 5
\]

 

We already have \( b \cdot c = 3 \) and \( a \cdot c = 4 \). To satisfy this equation, we explore possible values of \( \lfloor c \rfloor \).

 

Let’s assume \( \lfloor c \rfloor = 1 \). Substituting into Equation (3):

\[
a \cdot b \cdot 1 = 5
\]

 

So:

 

\[
a \cdot b = 5 \tag{6}
\]

 

#### Step 4: Solve the system of equations

Now, we have three equations:

 

1. \( b \cdot c = 3 \)

2. \( a \cdot c = 4 \)

3. \( a \cdot b = 5 \)

 

We can solve this system step by step. First, solve for \( c \) from Equation (5):

 

\[
c = \frac{4}{a}
\]

 

Substitute this into Equation (4):

 

\[
b \cdot \frac{4}{a} = 3
\]

 

Simplifying:

 

\[
b = \frac{3a}{4}
\]

 

Now substitute this into Equation (6):

 

\[
a \cdot \frac{3a}{4} = 5
\]

 

Simplifying:

 

\[
\frac{3a^2}{4} = 5
\]

 

Multiply both sides by 4:

 

\[
3a^2 = 20
\]

 

Solve for \( a^2 \):

 

\[
a^2 = \frac{20}{3}
\]

 

So:

 

\[
a = \sqrt{\frac{20}{3}} = \frac{2\sqrt{15}}{3}
\]

 

#### Step 5: Find the value of \( b \)

Now that we have \( a = \frac{2\sqrt{15}}{3} \), substitute this back into the expression for \( b \):

 

\[
b = \frac{3a}{4} = \frac{3 \times \frac{2\sqrt{15}}{3}}{4} = \frac{2\sqrt{15}}{4} = \frac{\sqrt{15}}{2}
\]

 

Thus, the value of \( b \) is:

\[
b = \frac{\sqrt{15}}{2}
\]