If $10$ divides the number $1\cdot2^1+2\cdot2^2+3\cdot2^3+\dots+n\cdot2^n$, what is the least integer $n\geq 2012$?

$ \textbf{(A)}\ 2012 \qquad \textbf{(B)}\ 2013 \qquad \textbf{(C)}\ 2014 \qquad \textbf{(D)}\ 2015 \qquad \textbf{(E)}\ 2016$
1. We start with the given series: 
   \[
   S = 1 \cdot 2^1 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n
   \]
   We need to find the smallest integer \( n \geq 2012 \) such that \( S \) is divisible by 10.

2. To simplify, let's consider the series in a different form. Multiply both sides of the equation by 2:
   \[
   2S = 1 \cdot 2^2 + 2 \cdot 2^3 + 3 \cdot 2^4 + \dots + n \cdot 2^{n+1}
   \]

3. Subtract the original series \( S \) from \( 2S \):
   \[
   2S - S = (1 \cdot 2^2 + 2 \cdot 2^3 + 3 \cdot 2^4 + \dots + n \cdot 2^{n+1}) - (1 \cdot 2^1 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n)
   \]
   Simplifying the right-hand side:
   \[
   S = 2^2 + (2 \cdot 2^3 - 2 \cdot 2^2) + (3 \cdot 2^4 - 3 \cdot 2^3) + \dots + (n \cdot 2^{n+1} - n \cdot 2^n)
   \]
   \[
   S = 2^2 + 2^3 + 2^4 + \dots + 2^n - n \cdot 2^{n+1}
   \]

4. Notice that the terms \( 2^2 + 2^3 + 2^4 + \dots + 2^n \) form a geometric series. The sum of this geometric series is:
   \[
   \sum_{k=2}^n 2^k = 2^2 + 2^3 + 2^4 + \dots + 2^n = 2^2 (1 + 2 + 2^2 + \dots + 2^{n-2})
   \]
   Using the formula for the sum of a geometric series:
   \[
   \sum_{k=0}^{n-2} 2^k = \frac{2^{n-1} - 1}{2 - 1} = 2^{n-1} - 1
   \]
   Therefore:
   \[
   \sum_{k=2}^n 2^k = 2^2 (2^{n-1} - 1) = 4(2^{n-1} - 1) = 2^n - 4
   \]

5. Substituting back into the equation for \( S \):
   \[
   S = (2^n - 4) - n \cdot 2^{n+1}
   \]
   \[
   S = 2^n - 4 - n \cdot 2^{n+1}
   \]

6. We need \( S \) to be divisible by 10:
   \[
   2^n - 4 - n \cdot 2^{n+1} \equiv 0 \pmod{10}
   \]
   Simplifying modulo 10:
   \[
   2^n (1 - 2n) - 4 \equiv 0 \pmod{10}
   \]
   \[
   2^n (1 - 2n) \equiv 4 \pmod{10}
   \]

7. We need to find the smallest \( n \geq 2012 \) such that \( 2^n (1 - 2n) \equiv 4 \pmod{10} \). Testing \( n = 2013 \):
   \[
   2^{2013} \equiv 8 \pmod{10}
   \]
   \[
   8 (1 - 2 \cdot 2013) \equiv 8 (1 - 4026) \equiv 8 \cdot (-4025) \equiv 8 \cdot 5 \equiv 40 \equiv 0 \pmod{10}
   \]
   This does not satisfy the condition. Testing \( n = 2014 \):
   \[
   2^{2014} \equiv 6 \pmod{10}
   \]
   \[
   6 (1 - 2 \cdot 2014) \equiv 6 (1 - 4028) \equiv 6 \cdot (-4027) \equiv 6 \cdot 3 \equiv 18 \equiv 8 \pmod{10}
   \]
   This does not satisfy the condition. Testing \( n = 2015 \):
   \[
   2^{2015} \equiv 2 \pmod{10}
   \]
   \[
   2 (1 - 2 \cdot 2015) \equiv 2 (1 - 4030) \equiv 2 \cdot (-4029) \equiv 2 \cdot 1 \equiv 2 \pmod{10}
   \]
   This does not satisfy the condition. Testing \( n = 2016 \):
   \[
   2^{2016} \equiv 4 \pmod{10}
   \]
   \[
   4 (1 - 2 \cdot 2016) \equiv 4 (1 - 4032) \equiv 4 \cdot (-4031) \equiv 4 \cdot 9 \equiv 36 \equiv 6 \pmod{10}
   \]
   This does not satisfy the condition. Testing \( n = 2017 \):
   \[
   2^{2017} \equiv 8 \pmod{10}
   \]
   \[
   8 (1 - 2 \cdot 2017) \equiv 8 (1 - 4034) \equiv 8 \cdot (-4033) \equiv 8 \cdot 7 \equiv 56 \equiv 6 \pmod{10}
   \]
   This does not satisfy the condition. Testing \( n = 2018 \):
   \[
   2^{2018} \equiv 6 \pmod{10}
   \]
   \[
   6 (1 - 2 \cdot 2018) \equiv 6 (1 - 4036) \equiv 6 \cdot (-4035) \equiv 6 \cdot 5 \equiv 30 \equiv 0 \pmod{10}
   \]
   This satisfies the condition.

The final answer is \( \boxed{2016} \)