What is the least integer $n\geq 100$ such that $77$ divides $1+2+2^2+2^3+\dots + 2^n$ ? 

$ \textbf{(A)}\ 101
\qquad\textbf{(B)}\ 105 
\qquad\textbf{(C)}\ 111
\qquad\textbf{(D)}\ 119
\qquad\textbf{(E)}\ \text{None}
$
1. We start with the given series sum \(1 + 2 + 2^2 + 2^3 + \dots + 2^n\). This is a geometric series with the first term \(a = 1\) and common ratio \(r = 2\). The sum of the first \(n+1\) terms of a geometric series is given by:
   \[
   S = \frac{r^{n+1} - 1}{r - 1}
   \]
   Substituting \(r = 2\), we get:
   \[
   S = \frac{2^{n+1} - 1}{2 - 1} = 2^{n+1} - 1
   \]

2. We need \(2^{n+1} - 1\) to be divisible by \(77\). Since \(77 = 7 \times 11\), we need \(2^{n+1} - 1\) to be divisible by both \(7\) and \(11\).

3. First, consider the divisibility by \(7\). By Fermat's Little Theorem, for any integer \(a\) and a prime \(p\), \(a^{p-1} \equiv 1 \pmod{p}\). Here, \(p = 7\), so:
   \[
   2^6 \equiv 1 \pmod{7}
   \]
   This implies that the order of \(2\) modulo \(7\) is \(6\). Therefore, \(2^{n+1} \equiv 1 \pmod{7}\) if and only if \(n+1\) is a multiple of \(3\) (since \(2^3 \equiv 1 \pmod{7}\)).

4. Next, consider the divisibility by \(11\). Again, by Fermat's Little Theorem:
   \[
   2^{10} \equiv 1 \pmod{11}
   \]
   This implies that the order of \(2\) modulo \(11\) is \(10\). Therefore, \(2^{n+1} \equiv 1 \pmod{11}\) if and only if \(n+1\) is a multiple of \(10\).

5. Combining these two conditions, we need \(n+1\) to be a multiple of both \(3\) and \(10\). The least common multiple of \(3\) and \(10\) is \(30\). Therefore, \(n+1\) must be a multiple of \(30\).

6. We are asked to find the least integer \(n \geq 100\) such that \(n+1\) is a multiple of \(30\). The smallest multiple of \(30\) that is greater than \(100\) is \(120\). Thus:
   \[
   n+1 = 120 \implies n = 119
   \]

Conclusion:
The least integer \(n \geq 100\) such that \(77\) divides \(1 + 2 + 2^2 + 2^3 + \dots + 2^n\) is \(\boxed{119}\).