I have had a problem with this concept all my life so I thought I would reach out to the experts for help!
Here is the problem statement:
Quote: "Consider how much work is required to multiply two -digit numbers using the usual grade-school method. There are two phases to working out the product: multiplication and addition.
First, multiply the first number by each of the digits of the second number. For each digit in the second number this requires $n$ basic operations (multiplication of single digits) plus perhaps some "carries", so say a total of $2n$ operations for each digit in the second number. This means that the multiplication phase requires basic $n(2n)$ operations.
The addition phase requires repeatedly adding n digit numbers together a total of $(n-1)$ times. If each addition requires at most $2n$ operations (including the carries), and there are $(n-1)$ additions that must be made, it comes to a total of $(2n)(n-1)$ operations in the addition phase.
Adding these totals up gives about $4n^2$ total operations. Thus, the total number of basic operations that must be performed in multiplying two n digit numbers is in $O(n^2)$ (since the constant coefficient does not matter)."
Why is it $(n-1)$ and $2n?$