Multiplying Larger Numbers

Short Course 

This is a short course (slide show) on how to multiply larger numbers.  Just click on the arrow on the right to start the show. 

Multiplying Larger Numbers

The process for calculating the final product involves multiplying each digit of multiplicand by each digit of the multiplier.  The process is the same process that we used in the previous chapter.  The only difference is that we start with the left most digit of the multiplier (the most significant digit) and repeat the process for each digit.

It is important to remember these two principles:

• • • Multiplying 2 digits always results in a 2-digit result.  If we multiply 3 time 3 the product (result) is 09.   If we multiply 4 times 4, the product is 16.

    • Multiplying two numbers, using the Double-A method, is always performed from left to right. While this method is different from what you learned in school, it produces the answer more quickly and preserves the place value of the result. 

Now, let’s review the process of calculating the product using larger numbers. The setup of the calculation is the same as in the previous chapter.  Remember, in this system, the first number is called the multiplicand and the second number is the multiplier.  The multiplicand is entered on the left side of the top abacus. The multiplier is entered in the right side of the top abacus. The result (or product) is entered and added in the bottom abacus.  If you switch the two numbers, you will still get the same result. 

In the previous chapter, we reviewed the idea that each digit in a number has a place value. You will recall that each digit in a number has a place value.  The value of each digit increases as we move from right to left.  We used the example, 43,567, where we have 4 ten thousands, 3 thousands, 5 hundreds, 6 tens and 7 ones. The most significant digit (the digit with the highest value) is the 4 in the ten thousands column. 

When we multiply two numbers, we multiply the most significant digits first. Moving from left to right, we multiply each digit of the multiplicand by each digit of the multiplier, adding each product in the bottom abacus.  This is the same process we used in the previous chapter with one added step.  The added step is that we need to repeat the multiplication process for each digit in the multiplier shifting one column to the right until we have multiplied all the digits in the multiplicand by each digit of the multiplier.

Let’s see how this works in an example, 43,567 times 35. Each digit of the multiplicand (with its place value) is multiplied by each digit of the multiplier:

   (40000 * 30)  =   1200000   +   Partial sum  =   1,200,000

       (3000 * 30) =     090000   +   Partial sum   =   1,290,000

             (500 * 30) =        15000   +   Partial sum    =   1,305,000

  (60 * 30) =           1800   +   Partial sum    =   1,306,800

     (7 * 30) =              210         Final sum       =   1,307,010

When we use our top and bottom abacus in the Double-A method, the starting column for the partial product in the bottom abacus is determined by adding the number of digits we are multiplying and the number of digits in the multiplier.  In this example, the multiplication of each digit is:

   (40000 * 30) =   1200000   +   Partial sum =   1,200,000   Starting column is  5 + 2 = 7

           (3000 * 30) =    090000   +   Partial sum =   1,290,000   Starting column is   4 + 2 = 6

           (500 * 30) =        15000   +   Partial sum =   1,305,000    Starting column is   3 + 2 = 5

 (60 * 30) =       1800   +   Partial sum =   1,306,800   Starting column is   2 + 2 = 4

    (7 * 30) =               210       Partial sum  =   1,307,010    Starting column is   1 + 2 = 3

Now we need to repeat the process for the second digit in the multiplier, 5, shifting one column to the right.

   (40000 * 5) =   200000   +   Partial sum =  1,507,010   Starting column is 5 + 1 = 6

       (3000 * 5) =   15000    +   Partial sum =  1,522,010   Starting column is 4 + 1 = 5

           (500 * 5) =     2500    +   Partial sum =   1,524,510   Starting column is 3 + 1 = 4

(60 * 5) =     300    +   Partial sum =   1,524,810   Starting column is 2 + 1 = 3

  (7 * 5) =                 35         Final sum   =    1,524,845   Starting column is 1 + 1 = 2

Examples

Now, let’s look at some more examples of this process. In these examples, the multiplicand will have many digits, but the multiplier will always be a single digit.  Click on the link to show the steps to calculate the answer. 

• Example: 61 times 84

• Example: 36 times 49

• Example: 25 times 204

• Example: 257 times 324

• Example: 7,288 times 448

• Example: 6,542 times 3,724

Build Your Skill

Now you are ready to try some calculations on your own. Click on problems and their solutions to test your skill.