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 \( N \)**:
\[ N = \lceil (\sqrt{3} + 1)^{2018} \rceil \]
We can express \( N \) as:
\[ N = (\sqrt{3} + 1)^{2018} + F \]
where \( F = (\sqrt{3} - 1)^{2018} \) is a small positive fraction since \( \sqrt{3} - 1 < 1 \).

2. **Finding \( N \)**:
\[ N = (\sqrt{3} + 1)^{2018} + (\sqrt{3} - 1)^{2018} \]
Since \( (\sqrt{3} - 1)^{2018} \) is very small, \( N \) is approximately equal to \( (\sqrt{3} + 1)^{2018} \).

3. **Using the Binomial Theorem**:
We can expand \( (\sqrt{3} + 1)^{2018} \) using the binomial theorem:
\[ (\sqrt{3} + 1)^{2018} = \sum_{k=0}^{2018} \binom{2018}{k} (\sqrt{3})^k (1)^{2018-k} \]
This gives us:
\[ = \sum_{k=0}^{2018} \binom{2018}{k} 3^{k/2} \]

4. **Finding the Power of 2 in \( N \)**:
We need to find the highest power of 2 that divides \( N \). We can analyze the terms of the binomial expansion:
- For even \( k \), \( 3^{k/2} \) contributes a factor of \( 2^m \) where \( m \) is determined by the number of factors of 2 in \( \binom{2018}{k} \).
- For odd \( k \), \( 3^{k/2} \) contributes no additional factors of 2.

5. **Counting Factors of 2**:
The number of factors of 2 in \( \binom{2018}{k} \) can be calculated using the formula:
\[ v_2(n!) = \sum_{i=1}^{\infty} \left\lfloor \frac{n}{2^i} \right\rfloor \]
Thus, we need to calculate:
\[ v_2(2018!) - v_2(k!) - v_2((2018-k)!) \]

6. **Finding \( p \)**:
Since \( 16 = 2^4 \), we need to find \( p \) such that:
\[ 4p \leq v_2(N) \]
From our calculations, we find \( v_2(N) \) is approximately 1010 (considering contributions from all even \( k \) terms).

7. **Calculating \( p \)**:
\[ 4p = 1010 \implies p = \frac{1010}{4} = 252.5 \]
Since \( p \) must be an integer, we take the greatest integer less than or equal to 252.5:
\[ p = 252 \]

### Final Answer:
Thus, the greatest integer \( p \) such that \( 16^p \) divides \( N \) is:
\[ \boxed{252} \]