How many of the coefficients of $(x+1)^{65}$ cannot be divisible by $65$?

$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{None}$
To determine how many coefficients of $(x+1)^{65}$ are not divisible by $65$, we need to analyze the binomial coefficients $\binom{65}{k}$ for $k = 0, 1, 2, \ldots, 65$ and check their divisibility by $65$. Since $65 = 5 \times 13$, we need to check the divisibility by both $5$ and $13$.

We will use Lucas' Theorem, which states that a binomial coefficient $\binom{n}{m}$ is divisible by a prime $p$ if and only if at least one of the base $p$ digits of $m$ is greater than the corresponding digit of $n$.

1. **Convert $65$ to base $5$ and base $13$:**
   - In base $5$: $65 = 2 \cdot 5^2 + 3 \cdot 5^1 + 0 \cdot 5^0 = (230)_5$
   - In base $13$: $65 = 5 \cdot 13^1 + 0 \cdot 13^0 = (50)_{13}$

2. **Identify $k$ values for which $\binom{65}{k}$ is not divisible by $5$:**
   - For $\binom{65}{k}$ to be not divisible by $5$, the base $5$ digits of $k$ must be less than or equal to the corresponding digits of $65$ in base $5$.
   - $65$ in base $5$ is $(230)_5$. Therefore, $k$ can be written as $(abc)_5$ where $0 \leq a \leq 2$, $0 \leq b \leq 3$, and $0 \leq c \leq 0$.
   - The possible values of $k$ in base $5$ are: $(000)_5, (001)_5, (002)_5, (010)_5, (011)_5, (012)_5, (020)_5, (021)_5, (022)_5, (030)_5, (031)_5, (032)_5, (100)_5, (101)_5, (102)_5, (110)_5, (111)_5, (112)_5, (120)_5, (121)_5, (122)_5, (130)_5, (131)_5, (132)_5, (200)_5, (201)_5, (202)_5, (210)_5, (211)_5, (212)_5, (220)_5, (221)_5, (222)_5, (230)_5$.
   - There are $3 \times 4 \times 1 = 12$ such values.

3. **Identify $k$ values for which $\binom{65}{k}$ is not divisible by $13$:**
   - For $\binom{65}{k}$ to be not divisible by $13$, the base $13$ digits of $k$ must be less than or equal to the corresponding digits of $65$ in base $13$.
   - $65$ in base $13$ is $(50)_{13}$. Therefore, $k$ can be written as $(ab)_{13}$ where $0 \leq a \leq 5$ and $0 \leq b \leq 0$.
   - The possible values of $k$ in base $13$ are: $(00)_{13}, (01)_{13}, (02)_{13}, (03)_{13}, (04)_{13}, (05)_{13}, (10)_{13}, (11)_{13}, (12)_{13}, (13)_{13}, (14)_{13}, (15)_{13}, (20)_{13}, (21)_{13}, (22)_{13}, (23)_{13}, (24)_{13}, (25)_{13}, (30)_{13}, (31)_{13}, (32)_{13}, (33)_{13}, (34)_{13}, (35)_{13}, (40)_{13}, (41)_{13}, (42)_{13}, (43)_{13}, (44)_{13}, (45)_{13}, (50)_{13}$.
   - There are $6 \times 1 = 6$ such values.

4. **Combine the results:**
   - We need to find the number of $k$ values for which $\binom{65}{k}$ is not divisible by either $5$ or $13$.
   - Using the principle of inclusion-exclusion, we find the number of $k$ values that are not divisible by $65$ (i.e., not divisible by both $5$ and $13$):
     \[
     \text{Total} = 12 + 6 - 2 = 16
     \]
   - Therefore, there are $16$ coefficients of $(x+1)^{65}$ that are not divisible by $65$.

The final answer is $\boxed{16}$