20*100 is the same as 100*20

First I shall derive an 2 by 3 algorithm and then show the 3 by 2 algorithm is the same. Suppose we have two numbers 41 and 247. We have many ways to represent these numbers. We could also represent them both like this:

(4*10) + (1*1) = 41 and (2*100) + (4*10) + (7*1) = 247

2 by 3 test:

((4*10) + 1*1)) * ((2*100) +(4*10) + (7*1)) =

(4*2*1000) + (4*4*100) + (4*7*10*1) + (1*2*100)+(1*4*10)+(1*7*1) =

(8*1000) + (16*100) + (28*10) + (2*100) + (4*10) + (7*1) =

8000 + 1600 + 280 + 200 + 40 + 7 =

9600 + 480 + 47 = 10127

3 by 2 test:

((2*100) +(4*10) + (7*1)) * ((4*10) + 1*1)) =

(2*4*100*10) + (2*1*100*1)+ (4*10*4*10) + (4*10*1*1) + (7*1*4*10) + (7*1*1*1) =

(8*1000) + (2*100) + (16*100) + (4*10) + (28*10) + 7 =

8000 + 200 + 1600 + 40 + 280 + 7 = 10127

Now lets derive the 2x3 formula. If we substitute the letters for the following set of numbers in the original problem above:

a = 4; b = 1; c = 2; d = 4; e=7

we could represent each number like this:

(a*10) + (b*1) = 41 and (c*100) + (d*10) + (e*1) = 247

But since we are solving for any number we don't have to worry about the numbers on the right hand side at the moment. Also note by convention, 10a is considered to be 10*a.

Now lets multiply the general case together

((a*10) + (b*1)) * ((c*100)+(d*10)+(e*1)) =

(10*a*c*100) + (a*10*d*10) + (a*10+e*1) + (b*1*c*100) + (b*1*d*10) + (b*1*e*1) =

1000ac + 100ad + 10ae + 100bc + 10bd + be = The answer to our problem.

Just for clarity we will now derive the 3 by 2 formula, again if we make the following substitutions:

a = 4; b = 1; c = 2; d = 4; e=7

we could represent each number like this:

(c*100) + (d*10) + (e*1) = 247 and (a*10) + (b*1) = 41

Next we multiply them together:

((c*100) + (d*10) + (e*1)) * ((a*10) + (b*1)) =

(c*100*a*10) + (c*100*b*1) + (d*10*a*10) + (d*10*b*1) + (e*1*a*10) + (e*1*b*1) =

1000ac + 100bc + 100ad + 10bd + 10ae + be = Notice this is the same answer as for the 2 by 3 formula

Since we have discovered the formula is the same for both the 2 by 3 case and the 3 by 2 case, the patterns for these are therefore both the same. This is a nice thing to have and it makes solutions by pattern solving a little easier. Now that we have formulas, its interesting to think of the problem like this: (generic a,b,c,d,e case)

The original numbers are arranged above the line. The formula generates the grouping below the line and shows them generally aligned according to their units.

Notice in our generic formula, the first digits start in the 100's column. The last digits start in the one's column.

There are two middle sets. each is comprised of two cross groupings as seen here

Each vertical grouping is added together. For example, in the left middle grouping, you multiply a*d and then multiply b*c and then add the two results together. I find the pictorial way of viewing the numbers with the highlighted groupings a much easier way to solve problems, then just memorizing some formula which you will probably forget . The eventual goal is to develop a pattern capable of handling a number of 'n' digits multiplied by a second number of 'm' digits.