What makes a number binomial? Identity 4

The equation can be written as $\sum _{ n=0 }^{ 10 }{ n\left( \begin{matrix} 10 \\ n \end{matrix} \right) }$

By using identity $\left( \begin{matrix} n \\ k \end{matrix} \right) =\left( \begin{matrix} n \\ n-k \end{matrix} \right)$ , we can get that $n\left( \begin{matrix} 10 \\ n \end{matrix} \right) +\left( 10-n \right) \left( \begin{matrix} 10 \\ 10-n \end{matrix} \right) =n\left( \begin{matrix} 10 \\ n \end{matrix} \right) +\left( 10-n \right) \left( \begin{matrix} 10 \\ n \end{matrix} \right) =10\left( \begin{matrix} 10 \\ n \end{matrix} \right) =5\left( \left( \begin{matrix} 10 \\ n \end{matrix} \right) +\left( \begin{matrix} 10 \\ 10-n \end{matrix} \right) \right)$

Then, $\sum _{ n=0 }^{ 10 }{ n\left( \begin{matrix} 10 \\ n \end{matrix} \right) }$ can be written as $5 \times \sum _{ n=0 }^{ 10 }{ \left( \begin{matrix} 10 \\ n \end{matrix} \right) }$ which result is $5 \times 2^{10}$ and $A = 5$

So, the last digit of $A$ is $\boxed{5}$