Complexity of long multiplication

Again, let's take a look at long multiplication. Suppose we are multiplying two numbers, each of which has $n$ digits.

     123
     521
    ====
     123
    246        
   615
   ======
   64083 

As you can see, we multiply each digit with each other digit, for a total of $n^2$ multiplications of two digits together (multiplying two digits together is a constant-time operation.)

Each product is then added onto a sum once (when computing the sum $64,083$). That means that in total, we perform approximately $n^2$ additions.

So in total, we perform $\mathcal{O}(n^2)$ operations.