Multiplication using vedic mathematics

INTRODUCTION:

There are a wide range of multiplication techniques, the one perhaps most familiar to the majority of people is the classic long multiplication algorithm, e.g.

        23958233
            5830 ×
    ------------
        00000000 (= 23,958,233 ×     0)
       71874699  (= 23,958,233 ×    30)
     191665864   (= 23,958,233 ×   800)
    119791165    (= 23,958,233 × 5,000)
    ------------
    139676498390 (= 139,676,498,390)

While this algorithm works for any pair of numbers, it is long winded, requires many intermediate stages, and requires you to record the results of each of the intermediate stages so you can sum them at the end to produce the final answer. However, when the numbers to be multiplied fall into certain categories, short-cuts can be used to avoid much of the work involved in long multiplication. There are many of these 'special cases' some of them allow seemingly difficult multiplications to be completed mentally, literally allowing you to just write down the answer.

Special Cases of Multiplication[edit]

There are a number of simple techniques that can be used when multiplying by certain numbers. They are very useful in their own right but can be even more useful when combined with other techniques where they can facilitate the solution of more difficult problems.

Multiplying by 11[edit]

To multiply any number by 11 do the following:
Working from right to left

1. Write the rightmost digit of the starting number down.
2. Add each pair of digits and write the results down, (carrying digits where necessary right to left).
3. Finally write down the left most digit (adding any final carry if necessary).

It's as simple as that, e.g.

• Multiply 712x11

${\displaystyle {\begin{matrix}&7&&1&&2\\&\swarrow \searrow &+&\swarrow \searrow &+&\swarrow \searrow \\7&&8&&3&&2\end{matrix}}}$

712x11=7832

The reason for working from right to left instead of the more usual left to right is so any carries can be added in. e.g.

• Multiply 8738x11

${\displaystyle {\begin{matrix}&8&&7&&3&&8\\&\swarrow \searrow &+&\swarrow \searrow &+&\swarrow \searrow &+&\swarrow \searrow \\9&\leftarrow _{1}&6&\leftarrow _{1}&1&\leftarrow _{1}&1&&8\end{matrix}}}$

8738x11=96118

Multiplying by 15[edit]

Multiplying by 15 can be broken down into a multiplication by 10 plus a multiplication by 5. Multiplying by 10 is just a matter of adding a 0 on the end of a number, multiplying by 5 is half of a multiplication by 10 as above e.g.

15 x 33 = [10 x 33] + [5 x 33] = 330 + [(10 x 33) / 2] = 330 + [330 / 2] = 330 + 165 = 495

This is easier to do in your mind than it is to write down! e.g. to multiply 15 x 27 you first multiply 27 x 10 giving 270, then 'add half again', i.e. half of 270 is 135, add this to 270 to get 405 and that's your answer.

Multiplying two single digit numbers[edit]

Although most people have memorised multiplication tables from 1x1 to 10x10, one of the Vedic Sutras, (Vertically and Crosswise), allows you to multiply any pair of single digit numbers without using anything higher than the 5x multiplication table. While this may not be particularly useful, the algorithm is a good introduction to some of the ideas behind the Vedic techniques and so is worth taking the time to learn and understand as the basic idea is expanded upon later. Because of this I will go through the procedure in more detail than it probably deserves. The technique is as follows:

1. If either of the numbers are below 6 then just recall the answer from memory (you need to know your multiplication tables up to 5x, hopefully this shouldn't be too much of a problem!). If instead both numbers are above 5 then continue.
2. Write, (or imagine), the two single digit numbers one above the other with an answer line below.
3. Subtract each number from 10 and place the result to the right of the original number.
4. Vertically: Multiply the two numbers on the right (the answers to the subtractions in the previous step) and place the answer underneath them on the answer line, (this is the first part of the answer). Since the original numbers were above 5, these numbers will always be below 5 (because the original numbers were subtracted from 10) so you won't need anything above the 4x multiplication table. If the answer to the multiplication is 10 or more, just place the right-most digit on the answer line and remember to carry the other digit to the next step.
5. Crosswise: Select one of the original numbers, (it doesn't matter which one, the answer will be the same), and subtract the number diagonally opposite it. If there was no carry from the previous step just place the result on the answer line below the original numbers, if there was a carry add this to the result before you place it on the answer line.

That's it, the number on the answer line is the final answer. The technique is very simple but looks more complicated than it actually is when written down step by step. The following examples should clarify the procedure.

• Multiply 8 x 6

Following steps 1 to 3 above, write the numbers down one above the other, subtract each from 10, and write the answer to the right of each number:

${\displaystyle {\begin{matrix}8&\longrightarrow &2&(10-8=2)\\\\6&\longrightarrow &4&(10-6=4)\\\hline \quad \end{matrix}}}$

Now, following step 4 (Vertically) multiply the two numbers on the right and write down the answer.

${\displaystyle {\begin{matrix}8&&2\\&&\downarrow \\6&&4\\\hline &&8&(2\times 4=8)\end{matrix}}}$

Finally, following step 5 (Crosswise) Subtract along any diagonal (the answer will be the same either way) and write down the answer.

${\displaystyle {\begin{matrix}&8&&2\\&&\swarrow &\\&6&&4\\\hline (6-2=4)&4&&8\end{matrix}}}$

So 8 x 6 is 48 as expected, but as you can see from the above sequence, to work this out you only needed to know how to subtract small numbers and multiply 2 x 4.

• Multiply 7 x 8

${\displaystyle {\begin{matrix}7&\longrightarrow &3\\\\8&\longrightarrow &2\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}7&&3\\&&\downarrow \\8&&2\\\hline &&6\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}7&&3\\&\swarrow &\\8&&2\\\hline 5&&6\end{matrix}}}$

Giving an answer of 56 as expected.
The next example involves a carry between the last two stages.

• Multiply 7 x 6

${\displaystyle {\begin{matrix}7&\longrightarrow &3\\\\6&\longrightarrow &4\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}7&&3\\&&\downarrow \\6&_{1}&4\\\hline &&2\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}7&&3\\&\swarrow &\\6&_{1}&4\\\hline 3&&2\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}7&&3\\&&\\6&&4\\\hline 4&&2\end{matrix}}}$

As you can see the multiplication in the second (Vertically) stage (3x4) results in 12, the 2 is written down and the 1 is carried. The subtraction in the Crosswise stage (6-3) results in 3, but you have to add the carried 1 from the previous step resulting in 4. The final answer is 42.

This may all seem like a lot of work for doing simple multiplications that you can just recall from memory, but the important thing here is to learn the technique as it is expanded upon later. If you are interested in improving your mental arithmetic skills, learn to do this procedure in your head by visualising the numbers laid out as above. Before going on to the next technique, try it now; close your eyes and multiply 6x6 using the procedure above...
You should have imagined the following:

• Multiply 6 x 6

${\displaystyle {\begin{matrix}6&\longrightarrow &4\\\\6&\longrightarrow &4\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}6&&4\\&&\downarrow \\6&_{1}&4\\\hline &&6&\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}6&&4\\&\swarrow &\\6&_{1}&4\\\hline 2&&6\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}6&&4\\&&\\6&&4\\\hline 3&&6&\end{matrix}}}$

Reference:https://en.wikibooks.org/wiki/Vedic_Mathematics/Techniques/Multiplication