If `0 lt x lt 1000 and [(x)/(2)] +[(x)/(3)]+[(x)/(5)] =(31)/(30)x`, where [x] is the greatest integer less than or equal to x, the number of possible values of x is

Options

A. 34
B. 32
C. 33
D. None of these

Answer

To solve the equation given in the question, we will follow these steps:

### Step 1: Understand the equation
The equation we have is:
\[ \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor + \left\lfloor \frac{x}{5} \right\rfloor = \frac{31}{30} x \]
where \(\left\lfloor x \right\rfloor\) denotes the greatest integer less than or equal to \(x\).

### Step 2: Determine the conditions for \(x\)
Given that \(0 < x < 1000\), we need to ensure that the right-hand side of the equation is an integer. For \(\frac{31}{30} x\) to be an integer, \(x\) must be a multiple of 30. Thus, we can express \(x\) as:
\[ x = 30p \]
where \(p\) is an integer.

### Step 3: Substitute \(x\) in the equation
Now substituting \(x = 30p\) into the left-hand side of the equation:
\[ \left\lfloor \frac{30p}{2} \right\rfloor + \left\lfloor \frac{30p}{3} \right\rfloor + \left\lfloor \frac{30p}{5} \right\rfloor \]
Calculating each term:
\[ \left\lfloor \frac{30p}{2} \right\rfloor = \left\lfloor 15p \right\rfloor = 15p \]
\[ \left\lfloor \frac{30p}{3} \right\rfloor = \left\lfloor 10p \right\rfloor = 10p \]
\[ \left\lfloor \frac{30p}{5} \right\rfloor = \left\lfloor 6p \right\rfloor = 6p \]
Adding these together gives:
\[ 15p + 10p + 6p = 31p \]

### Step 4: Set the left-hand side equal to the right-hand side
Now we set the left-hand side equal to the right-hand side:
\[ 31p = \frac{31}{30} (30p) \]
This simplifies to:
\[ 31p = 31p \]
This equality holds for all \(p\), confirming that our substitution is valid.

### Step 5: Determine the range for \(p\)
Since \(0 < x < 1000\), substituting \(x = 30p\) gives:
\[ 0 < 30p < 1000 \]
Dividing the entire inequality by 30:
\[ 0 < p < \frac{1000}{30} \approx 33.33 \]
Thus, \(p\) can take integer values from 1 to 33 (inclusive).

### Step 6: Count the possible values of \(p\)
The possible integer values for \(p\) are:
\[ 1, 2, 3, \ldots, 33 \]
This gives us a total of 33 possible values for \(p\).

### Final Answer
Thus, the number of possible values of \(x\) is:
\[ \boxed{33} \]