For $n \geq 1$, let $a_n$ be the number beginning with $n$ $9$'s followed by $744$; eg., $a_4=9999744$. Define $$f(n)=\text{max}\{m\in \mathbb{N} \mid2^m ~ \text{divides} ~ a_n \}$$, for $n\geq 1$. Find $f(1)+f(2)+f(3)+ \cdots + f(10)$.
1. Define \( a_n \) as the number beginning with \( n \) 9's followed by 744. For example, \( a_4 = 9999744 \).
2. We need to find \( f(n) = \text{max}\{m \in \mathbb{N} \mid 2^m \text{ divides } a_n \} \) for \( n \geq 1 \).
3. Note that \( a_n = 10^{n+3} - 2^8 \). This is because \( 10^{n+3} \) is a number with \( n+3 \) digits, all 0's, and subtracting 256 (which is \( 2^8 \)) gives us the desired number ending in 744.
4. We need to find the highest power of 2 that divides \( 10^{n+3} - 256 \). This can be done using the \( v_2 \) function, which gives the highest power of 2 dividing a number.

### Case 1: \( n \leq 5 \)
For \( n \leq 5 \), \( n+3 \leq 8 \). We can factor \( 2^{n+3} \) out of \( 10^{n+3} - 256 \):
\[ 10^{n+3} - 256 = 2^{n+3} \cdot (5^{n+3} - 2^{5-n}) \]
Since \( 5^{n+3} - 2^{5-n} \) is odd (as \( 5^{n+3} \) is odd and \( 2^{5-n} \) is even), the highest power of 2 dividing \( 10^{n+3} - 256 \) is \( 2^{n+3} \).
Thus, \( f(n) = n+3 \) for \( n \leq 5 \).

### Case 2: \( n \geq 6 \)
For \( n \geq 6 \), \( n+3 \geq 9 \). We can factor \( 2^8 \) out of \( 10^{n+3} - 256 \):
\[ 10^{n+3} - 256 = 2^8 \cdot (5^{n+3} \cdot 2^{n-5} - 1) \]
We need to find the highest power of 2 dividing \( 5^{n+3} \cdot 2^{n-5} - 1 \). For \( n \geq 6 \), \( 5^{n+3} \cdot 2^{n-5} - 1 \) is odd except for \( n = 8 \) where it is divisible by \( 2^5 \) (as \( 5^{11} \cdot 2^3 - 1 \) is divisible by \( 2^5 \)).
Thus, \( f(n) = 8 \) for \( n \geq 6 \) except \( n = 8 \), where \( f(8) = 13 \).

### Summarizing the values:
\[
\begin{align*}
f(1) &= 4 \\
f(2) &= 5 \\
f(3) &= 6 \\
f(4) &= 7 \\
f(5) &= 8 \\
f(6) &= 8 \\
f(7) &= 8 \\
f(8) &= 13 \\
f(9) &= 8 \\
f(10) &= 8 \\
\end{align*}
\]

### Summing these values:
\[
f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) = 4 + 5 + 6 + 7 + 8 + 8 + 8 + 13 + 8 + 8 = 75
\]

The final answer is \(\boxed{75}\)