Let N be the integer next above `(sqrt3+1)^(2018)`. The greatest integer ‘p’ such that `(16)^(p)` divides N is equal to:

Answer

To solve the problem, we need to find the greatest integer \( p \) such that \( 16^p \) divides \( N \), where \( N \) is the integer next above \( (\sqrt{3} + 1)^{2018} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with \( N = \lceil (\sqrt{3} + 1)^{2018} \rceil \). We will also consider \( (\sqrt{3} - 1)^{2018} \) because it will help us in simplifying our calculations.

2. **Approximation of Terms**:
We know that \( \sqrt{3} \approx 1.732 \), thus \( \sqrt{3} + 1 \approx 2.732 \) and \( \sqrt{3} - 1 \approx 0.732 \). Therefore, \( (\sqrt{3} - 1)^{2018} \) will be a very small positive number since \( 0.732 < 1 \).

3. **Using Binomial Theorem**:
We can express \( N \) as:
\[ N = (\sqrt{3} + 1)^{2018} + (\sqrt{3} - 1)^{2018} \]
The second term \( (\sqrt{3} - 1)^{2018} \) is negligible compared to the first term, so we can approximate:
\[ N \approx (\sqrt{3} + 1)^{2018} \]

4. **Finding the Power of 2**:
We need to find the power of 2 in \( N \). Using the binomial expansion:
\[ (\sqrt{3} + 1)^{2018} = \sum_{k=0}^{2018} \binom{2018}{k} (\sqrt{3})^k (1)^{2018-k} \]
The terms where \( k \) is even will contribute to the integer part, while the odd powers will cancel out when we add \( (\sqrt{3} - 1)^{2018} \).

5. **Calculating the Total Contribution**:
The terms with even \( k \) contribute positively, while the odd \( k \) terms will cancel out. The contribution of \( k \) being even can be expressed as:
\[ \sum_{j=0}^{1009} \binom{2018}{2j} 3^j \]
This sum will be an integer.

6. **Finding the Total Power of 2**:
We can find the power of 2 in \( N \) by analyzing the binomial coefficients:
\[ N \approx 2^{2018} \cdot \text{(some integer)} \]
To find the exact power of 2, we can use the fact that:
\[ \nu_2(N) = \nu_2\left( \sum_{j=0}^{1009} \binom{2018}{2j} 3^j \right) \]
The leading term will dominate, and we can estimate that:
\[ N \approx 2^{2018} \]

7. **Finding \( p \)**:
Since \( 16^p = 2^{4p} \), we set:
\[ 4p \leq 2018 \implies p \leq \frac{2018}{4} = 504.5 \]
Thus, the greatest integer \( p \) is:
\[ p = 504 \]

### Final Answer:
The greatest integer \( p \) such that \( 16^p \) divides \( N \) is \( \boxed{504} \).