Let `E = [(1)/(3) + (1)/(50)]+[(1)/(3)+(2)/(50)]+[(1)/(3)+(3)/(50)]+……..` upto `50` terms where `[.]` denotes the greatest integer terms then

Options

A. `E` is divisible by exactly `2` primes
B. `E` is prime
C. `E ge 30`
D. `E le 35`

Explanation

To solve the problem, we need to evaluate the expression \( E = \left[ \frac{1}{3} + \frac{1}{50} \right] + \left[ \frac{1}{3} + \frac{2}{50} \right] + \left[ \frac{1}{3} + \frac{3}{50} \right] + \ldots \) up to 50 terms, where \([.]\) denotes the greatest integer function.

### Step-by-step Solution:

1. **Understanding the Terms**:
Each term in the series can be expressed as:
\[ \left[ \frac{1}{3} + \frac{k}{50} \right] \quad \text{for } k = 1, 2, \ldots, 50 \]

2. **Finding the Value of Each Term**:
We need to determine when the expression \( \frac{1}{3} + \frac{k}{50} \) is less than 1 and when it is greater than or equal to 1.
- The critical point occurs when:
\[ \frac{1}{3} + \frac{k}{50} < 1 \]
Rearranging gives:
\[ \frac{k}{50} < 1 - \frac{1}{3} = \frac{2}{3} \]
Thus:
\[ k < \frac{2}{3} \times 50 = \frac{100}{3} \approx 33.33 \]
Therefore, \( k \) can take integer values from 1 to 33.

3. **Calculating the Greatest Integer Values**:
For \( k = 1, 2, \ldots, 33 \):
\[ \left[ \frac{1}{3} + \frac{k}{50} \right] = 0 \quad \text{(since it is less than 1)} \]
Thus, the contribution to \( E \) from these terms is:
\[ E_1 = 0 \text{ (from 33 terms)} \]

4. **Finding the Next Set of Terms**:
For \( k = 34, 35, \ldots, 50 \):
We check when:
\[ \frac{1}{3} + \frac{k}{50} \geq 1 \]
This occurs when:
\[ \frac{k}{50} \geq \frac{2}{3} \quad \Rightarrow \quad k \geq \frac{100}{3} \approx 33.33 \]
Thus, \( k \) can take values from 34 to 50 (17 terms).

5. **Calculating the Greatest Integer Values for Higher Terms**:
For \( k = 34, 35, \ldots, 50 \):
- The value of \( \frac{1}{3} + \frac{k}{50} \) will be in the range:
\[ \frac{1}{3} + \frac{34}{50} \text{ to } \frac{1}{3} + \frac{50}{50} \]
- The minimum value is:
\[ \frac{1}{3} + \frac{34}{50} = \frac{1}{3} + 0.68 = 1.01 \quad \Rightarrow \quad \left[ 1.01 \right] = 1 \]
- The maximum value is:
\[ \frac{1}{3} + 1 = \frac{4}{3} \quad \Rightarrow \quad \left[ \frac{4}{3} \right] = 1 \]
Therefore, for each of these 17 terms, we have:
\[ \left[ \frac{1}{3} + \frac{k}{50} \right] = 1 \]

6. **Summing Contributions**:
The total contribution from \( k = 34 \) to \( k = 50 \) is:
\[ E_2 = 1 \times 17 = 17 \]

7. **Final Calculation**:
Adding the contributions from both parts:
\[ E = E_1 + E_2 = 0 + 17 = 17 \]

### Final Answer:
\[ E = 17 \]