If `x,y in (0,30)` such that `[(x)/(3)]+[(3x)/(2)]+[(y)/(2)]+[(3y)/(4)]=(11)/(6)x+(5)/(4)y`
(where [x] denotes greatest integer `le x`), then number of ordered pairs (x,y) is

Options

A. 0
B. 2
C. 4
D. None of these

Explanation

To solve the equation

\[ \left\lfloor \frac{x}{3} \right\rfloor + \left\lfloor \frac{3x}{2} \right\rfloor + \left\lfloor \frac{y}{2} \right\rfloor + \left\lfloor \frac{3y}{4} \right\rfloor = \frac{11}{6}x + \frac{5}{4}y \]

where \( x, y \in (0, 30) \), we will analyze the left-hand side and right-hand side separately.

### Step 1: Analyze the greatest integer functions

The greatest integer function \( \left\lfloor z \right\rfloor \) gives the largest integer less than or equal to \( z \). We can express the left-hand side in terms of \( x \) and \( y \):

1. \( \left\lfloor \frac{x}{3} \right\rfloor \) can be written as \( \frac{x}{3} - \{ \frac{x}{3} \} \)
2. \( \left\lfloor \frac{3x}{2} \right\rfloor \) can be written as \( \frac{3x}{2} - \{ \frac{3x}{2} \} \)
3. \( \left\lfloor \frac{y}{2} \right\rfloor \) can be written as \( \frac{y}{2} - \{ \frac{y}{2} \} \)
4. \( \left\lfloor \frac{3y}{4} \right\rfloor \) can be written as \( \frac{3y}{4} - \{ \frac{3y}{4} \} \)

Combining these, we have:

\[ \left( \frac{x}{3} + \frac{3x}{2} + \frac{y}{2} + \frac{3y}{4} \right) - \left( \{ \frac{x}{3} \} + \{ \frac{3x}{2} \} + \{ \frac{y}{2} \} + \{ \frac{3y}{4} \} \right) \]

### Step 2: Simplify the left-hand side

Now we simplify the non-integer part:

\[ \frac{x}{3} + \frac{3x}{2} = \frac{2x + 9x}{6} = \frac{11x}{6} \]
\[ \frac{y}{2} + \frac{3y}{4} = \frac{2y + 3y}{4} = \frac{5y}{4} \]

Thus, the left-hand side becomes:

\[ \frac{11x}{6} + \frac{5y}{4} - \left( \{ \frac{x}{3} \} + \{ \frac{3x}{2} \} + \{ \frac{y}{2} \} + \{ \frac{3y}{4} \} \right) \]

### Step 3: Set the equation

Setting the left-hand side equal to the right-hand side gives:

\[ \frac{11x}{6} + \frac{5y}{4} - \left( \{ \frac{x}{3} \} + \{ \frac{3x}{2} \} + \{ \frac{y}{2} \} + \{ \frac{3y}{4} \} \right) = \frac{11}{6}x + \frac{5}{4}y \]

This implies:

\[ - \left( \{ \frac{x}{3} \} + \{ \frac{3x}{2} \} + \{ \frac{y}{2} \} + \{ \frac{3y}{4} \} \right) = 0 \]

### Step 4: Determine conditions for \( x \) and \( y \)

For the equation to hold true, each fractional part must equal zero:

1. \( \{ \frac{x}{3} \} = 0 \) implies \( x \) is a multiple of 3.
2. \( \{ \frac{3x}{2} \} = 0 \) implies \( x \) is a multiple of 2.
3. \( \{ \frac{y}{2} \} = 0 \) implies \( y \) is a multiple of 2.
4. \( \{ \frac{3y}{4} \} = 0 \) implies \( y \) is a multiple of 4.

Thus, \( x \) must be a multiple of 6 (LCM of 2 and 3) and \( y \) must be a multiple of 4.

### Step 5: Find valid multiples within the range

- For \( x \): The multiples of 6 in the range \( (0, 30) \) are \( 6, 12, 18, 24, 30 \) (5 values).
- For \( y \): The multiples of 4 in the range \( (0, 30) \) are \( 4, 8, 12, 16, 20, 24, 28 \) (7 values).

### Step 6: Calculate the number of ordered pairs

The total number of ordered pairs \( (x, y) \) is given by the product of the number of valid \( x \) values and \( y \) values:

\[ \text{Total pairs} = 5 \times 7 = 35 \]

### Final Answer

The number of ordered pairs \( (x, y) \) is \( \boxed{35} \).

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