Apply the idea
Set up the vertical algorithm: \begin{array}{c} &&2&5&0 \\ &\times &&8&0 \\ \hline &&&& \\ \hline \end{array}
First we will multiply 250 by 0.
Remember that all number that multiply to 0 is equal to 0. So we put 0 in the units, tens, and hundreds place.
\begin{array}{c} &&2&5&0 \\ &\times &&8&0 \\ \hline &&0&0&0 \\ \hline \end{array}
Now we will multiply 250 by the 8 in the tens place. We will write our answer underneath our previous answer.
Since we are multiplying by a number in the tens place we will place a 0 in the units place.
\begin{array}{c} &2&5&0 \\ \times& &8&0 \\ \hline &0&0&0 \\ &&& 0 \\ \hline \end{array}
8\times 0=0 so we put a 0 in the tens place:
\begin{array}{c} &2&5&0 \\ \times& &8&0 \\ \hline &0&0&0 \\ &&0& 0 \\ \hline \end{array}
8\times 5=40 so we put the 0 in the hundreds place and carry the 4 to the hundreds place.
\begin{array}{c} &{}^42&5&0 \\ \times& &8&0 \\ \hline &0&0&0 \\ &0&0& 0 \\ \hline \end{array}
8 \times 2=16 then add the carried 4 to get 20. Put the 0 in the thousands place and the 2 in the ten thousands place.
\begin{array}{c} &&& &{}^42&5&0 \\ &\times &&& &8&0 \\ \hline &&& &0&0&0 \\ &&2&0&0&0& 0 \\ \hline \end{array}
Add our two answers to get the final answer:
\begin{array}{c} &&& &{}^42&5&0 \\ &\times &&& &8&0 \\ \hline &&& &0&0&0 \\ &+&2&0&0&0& 0 \\ \hline &&2&0&0&0& 0 \\ \hline \end{array}