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Vedic Multiplication

Step 5: Vedic Multiplication Examples: 2-Digit by 2-Digit

Picture of Vedic Multiplication Examples: 2-Digit by 2-Digit
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In this step (and the following steps) we will attempt to clarify the vedic procedure through the use of several examples. Under each example, there will be a number of steps (indicated by numbers). Each step has a separate picture, so feel free to use those to follow along with what's going on.

Example 1: 42*21
1. Multiply the 2 highest digits (4 and 2), resulting in an 8.
2. For the next higher digit, cross multiply 4*1 (4) and 2*2 (4), and add together, producing the middle digit of 8.
3. For the lowest digit, multiply the 2 lowest digits (1*2) together, resulting in a 2.
4. Put all of the digits together to produce your answer (882)

A quick check on the calculator indicates our answer is indeed correct.

One thing that can and should be noted here is that the order in which you go through for the vedic process does not actually matter. So, we can similarly start with the lowest digit and work our way up to the highest digit.

Example 2: 67*23 (Starting with the lowest digit and working our way left)
1. Lowest Digit: Multiply the 2 lowest digits together (7*3), to get 21. Only one digit can be kept per place, so write a 1 and carry a 2 over to the next position.
2. Now, move over by 1 position on top, (3 will be multiplied by 6 now) and cross multiply pairs, decrementing the position on top while incrementing the position on the bottom. That is to say, multiply 3*6 (18) and 2*7 (14) and add them together (32). Write a 2, carry a 3 to the next position.

3. At this point, we have reached the end of the number, and thus incrementing the position on top is not possible. Thus, we simply increment the position on the bottom (2 will be multiplied by 6) and no cross multiplication will occur (as incrementing the bottom digit is not possible, we have reached the end of the number). So, 2*6 will be 12, so write a 2 and carry a 1.

4. Now that the multiplication has been completed, we have 2 numbers. The one on top is all of ones digits from the multiplications, and the one on the bottom is all of the carries that occurred. So, simply add these 2 numbers (221 and 1320) together, resulting in the answer (1541).

In the following steps, we will perform examples to expand the idea to higher orders of digits.
 
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igorzal5 years ago
Seems like a far easier method is the "SAT" approach...that they unfortunately teach only in prep for the SAT's rather than earlier, when it might do some good.

67x23 =
67*2 = 134
*10=1340
+67*3=201
=1541

or 67*25=
67*100=6700
/2=3350
/2=1675
purduecer (author)  igorzal5 years ago
The method you refer to I have frequently heard referred to as "cluster problems", and it can be useful for some cases. However, I am of the opinion that this method becomes less useful as you start applying it to higher numbers of digits (ex. 97623 * 85428). You quickly stretch the limits of your grey matter attempting to remember all of the little sub-problems you created to reach the final result, and you waste lots of time writing down all the little intermediate steps you had to take to get to the answer. For this reason, I recommend the vedic approach or at least the standard algorithm for multiplication, particularly if you intend to multiply something beyond 3-digits by 3-digits.
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