Multiplication using vedic Mathematics continues....

Multiplying two numbers between 89 to 100[edit]

The above technique actually works for any two numbers but it is only useful if it results in an easier process than traditional long multiplication. The key thing to remember is that with this technique you end up multiplying the subtracted numbers instead of the original numbers, i.e. it is only easier than normal multiplication if these subtracted numbers are smaller than the original numbers, hence the reason why we only used the above technique for numbers greater than 5, (since the subtracted numbers will then be 4 or smaller).

When the technique is extended to double digit numbers, you subtract each from 100 during the 'Vertically' stage instead of subtracting them from 10 so the technique is only easier if the result of this subtraction is small for one or both of the numbers, this is obviously the case when one or both of the numbers are close to 100. Take a look at the following example:

• Multiply 89 x 97

${\displaystyle {\begin{matrix}89&\longrightarrow &11&(100-89=11)\\\\97&\longrightarrow &3&(100-97=3)\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}89&&11\\&&\downarrow \\97&&3\\\hline &&33&(3\times 11=33)\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}&89&&11\\&&\nwarrow &\\&97&&3\\\hline (89-3=86)&86&&33\end{matrix}}}$

So 89x97=8633.
Now the power of the technique becomes clear. In the introduction I asked whether you wanted to be able to multiply 89x97 quickly in your head, you should now be able to see that this is actually quite easy. You first visualise both numbers one on top of the other, you then subtract each from 100 giving 11 and 3, mentally placing each result to the right of the original numbers. Next you multiply 11 and 3 together giving 33. (Note that since we are now dealing with double digit numbers, we don't carry unless the answer to this multiplication is 100 or more). We now have the last two digits of the answer (33), all we have to do now is subtract along either of the diagonals to get the first digits. We can pick any diagonal as they will always give the same answer but 89-3=86 is perhaps easier than 97-11=86. The final answer is the concatenation of the two parts giving 8633
If you practice this technique you will find you can do two digit multiplications without writing anything down. Try it now, multiply 95x93 in your head, try to visualise the procedure above.
You should have come up with the following:

• Multiply 95 x 93

${\displaystyle {\begin{matrix}95&\longrightarrow &5\\\\93&\longrightarrow &7\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}95&&5\\&&\downarrow \\93&&7\\\hline &&35\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}95&&5\\&\nwarrow &\\93&&7\\\hline 88&&35\end{matrix}}}$

95x93=8835

Remember that the reason why this is easier than normal multiplication is because you only have to multiply the results of the subtractions. This means that you can usually use it in situations where only one of the numbers is close to 100 because the multiplication will still be easy in this case. e.g.

• Multiply 97 x 69

${\displaystyle {\begin{matrix}97&\longrightarrow &3\\\\69&\longrightarrow &31\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}97&&3\\&&\downarrow \\69&&31\\\hline &&93\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}97&&3\\&\swarrow &\\69&&31\\\hline 66&&93\end{matrix}}}$

97x69=6693

• Multiply 96 x 88

${\displaystyle {\begin{matrix}96&\longrightarrow &4\\\\88&\longrightarrow &12\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}96&&4\\&&\downarrow \\88&&12\\\hline &&48\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}96&&4\\&\swarrow &\\88&&12\\\hline 84&&48\end{matrix}}}$

96x88=8448

The same technique works for numbers slightly over 100 except you now have to add during the Crosswise step. e.g.

• Multiply 105 x 107

${\displaystyle {\begin{matrix}105&\longrightarrow &5\\\\107&\longrightarrow &7\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}105&&5\\&&\downarrow \\107&&7\\\hline &&35\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}105&&5\\&\swarrow &\\107&&7\\\hline 112&&35\end{matrix}}}$

105x107=11235

There are a number of ways to remember the extension of the technique to numbers larger than 100. If you are familiar with the sign multiplication rules (i.e. -x-=+, -x+=-, etc.) then you don't have to alter the technique at all as you will understand that 100-105 = -5 and 100-107=-7, then (-7)x(-5)=35, and 107-(-5)=107+5=112.
If you are not comfortable with this then you can instead just reverse the initial subtractions when the original numbers are greater than 100 (or just remember that you need the difference between the original numbers and 100, i.e. 105-100 = 5 instead of 100-105 = -5), and then also remember that you have to change the Crosswise subtraction to an addition if the number you want to subtract came from an original number greater than 100.
Try to do the following examples in your head.

• Multiply 109 x 108

${\displaystyle {\begin{matrix}109&\longrightarrow &9\\\\108&\longrightarrow &8\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}109&&9\\&&\downarrow \\108&&8\\\hline &&72\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}109&&9\\&\swarrow &\\108&&8\\\hline 117&&72\end{matrix}}}$

109x108=11772

• Multiply 115 x 106

${\displaystyle {\begin{matrix}115&\longrightarrow &15\\\\106&\longrightarrow &6\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}115&&15\\&&\downarrow \\106&&6\\\hline &&90\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}115&&15\\&\swarrow &\\106&&6\\\hline 121&&90\end{matrix}}}$

115x106=12190

• Multiply 123 x 103

${\displaystyle {\begin{matrix}123&\longrightarrow &23\\\\103&\longrightarrow &3\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}123&&23\\&&\downarrow \\103&&3\\\hline &&69\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}123&&23\\&\swarrow &\\103&&3\\\hline 126&&69\end{matrix}}}$

123x103=12669

As before you need to carry the first digit of the Vertically multiplication if it is more than 2 digits long. e.g.

• Multiply 133 x 120

${\displaystyle {\begin{matrix}133&\longrightarrow &33\\\\120&\longrightarrow &20\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}133&&33\\&&\downarrow \\120&_{6}&20\\\hline &&60\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}133&&33\\&\swarrow &\\120&_{6}&20\\\hline 153&&60\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}133&&33\\&&\\120&&20\\\hline 159&&60\end{matrix}}}$

133x120=15960

Combining techniques[edit]

One of the key ideas of the Vedic system is that you can combine techniques to solve problems. You should look at problems in a flexible way and use the combination of techniques that best suits a particular problem, (and the way your own brain works!). It is perhaps a little early to be discussing this since we have only covered one major technique so far, but even at this stage it is possible to combine the Vertically and Crosswise multiplication technique with the special case multiplication techniques already described to handle the situation when both numbers are further away from 100, 1000, etc.

For example, using the 'by 10 and half again' rule to multiply by 15 lets you easily deal with numbers further away from 10, 100, 1000, etc., if one of the numbers is 15 away from your 'base' number e.g.

• Multiply 66 x 85

${\displaystyle {\begin{matrix}66&\longrightarrow &34\\\\85&\longrightarrow &15\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}66&&34\\&&\downarrow \\85&_{5}&15\\\hline &&10\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}66&&34\\&\nwarrow &\\85&_{5}&15\\\hline 51&&10\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}66&&34\\&&\\85&&15\\\hline 56&&10\end{matrix}}}$

66x85=5610
You can see here that the Vertically multiplication results in 510 (34x15 = 340 'and half again' = 340 + 170 = 510). 510 has 3 digits so we write down the last two digits (10) and carry the leading 5. We then do the Crosswise subtraction along the easiest diagonal (66-15 = 51) and add the carry (5) before writing down the final answer (56).

Extending the Multiplication Technique[edit]

The same Vertically wise technique described above works for any numbers but it is particularly useful for numbers near a power of 10, i.e. 10, 100, 1000, 10000, 100000, etc. As long as the initial subtraction results in numbers that are 'easier' to multiply it is a useful technique. e.g.

• Multiply 1232 x 1003

${\displaystyle {\begin{matrix}1232&\longrightarrow &232\\\\1003&\longrightarrow &3\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}1232&&232\\&&\downarrow \\1003&&3\\\hline &&696\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}1232&&232\\&\swarrow &\\1003&&3\\\hline 1235&&696\end{matrix}}}$

1232x1003=1235696
Since we are dealing with numbers near 1000 here we find the initial differences from 1000 instead of 100 and we only carry if the Vertically multiplication is 1000 or greater.

• Multiply 9960 x 9850

${\displaystyle {\begin{matrix}9960&\longrightarrow &40\\\\9850&\longrightarrow &150\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}9960&&40\\&&\downarrow \\9850&&150\\\hline &&6000\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}9960&&40\\&\swarrow &\\9850&&150\\\hline 9810&&6000\end{matrix}}}$

9960x9850=98106000
In this case the numbers are slightly below 10000 so we initially subtract from 10000, we also subtract in the Crosswise stage, and we only carry if the Vertically multiplication is 10000 or more. (In this particular case it would have perhaps been simpler to do 996x985 using the Vertically and Crosswise technique and then add two zeroes to the end of the answer.)

• Multiply 89684 x 99989

${\displaystyle {\begin{matrix}89684&\longrightarrow &10316\\\\99989&\longrightarrow &11\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}89684&&10316\\&&\downarrow \\99989&_{1}&11\\\hline &&13476\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}89684&&10316\\&\nwarrow &\\99989&_{1}&11\\\hline 89673&&13476\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}89684&&10316\\&&\\99989&&11\\\hline 89674&&13476\end{matrix}}}$

89684x99989=8967413476
Note the combination of techniques in the above example, i.e. the difference between 89684 and 100000 is easily derived using the special case technique for subtracting from a power of 10, (i.e. using the sutra All from 9 and the last from 10). The special case technique for multiplication by 11 is also used.

• Multiply 98688 x 99997

${\displaystyle {\begin{matrix}98688&\longrightarrow &1312\\\\99997&\longrightarrow &3\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}98688&&1312\\&&\downarrow \\99997&&3\\\hline &&3936\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}98688&&1312\\&\swarrow &\\99997&&3\\\hline 98685&&03936\end{matrix}}}$

98688x99997=9868503936
In this example the numbers are slightly less than 100000 so the initial subtraction is from 100000. The thing to watch out for here is that the result of the Vertically multiplication must be padded out to 5 digits by adding an extra zero on the left (03936 instead of 3936). This is also true generally, i.e. the number of digits in the result of the Verticallymultiplication must always be the same as the number of zeroes in the 'base' number (i.e. 100, 1000, 100000, etc.).

It is worth remembering how far we have come even at this early stage. Even if you are writing the calculations down, it is still much more efficient to do the multiplication above the Vedic way than using traditional long multiplication. e.g.

• Multiply 98688 x 99997 using long multiplication

           98688
           99997 ×
      ----------
          690816 (= 98688 ×     7)
         8881920 (= 98688 ×    90)
        88819200 (= 98688 ×   900)
       888192000 (= 98688 ×  9000)
      8881920000 (= 98688 × 90000)
      ----------
      9868503936

This multiplication technique can be extended further to cover cases where one number is slightly above a power of 10 and one slightly below the same power of 10. In this case it is advantageous to note whether the original numbers are greater or smaller than the 'base' power of 10 using + or - symbols accordingly. e.g.:

• Multiply 111 x 88

${\displaystyle {\begin{matrix}111&\longrightarrow &+11\\\\88&\longrightarrow &-12\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}111&&+11\\&&\downarrow \\88&&-12\\\hline &&-132\end{matrix}}}$

Note that the result of the Vertically multiplication is now negative because the signs of the two numbers you are multiplying are different. Additionally, since our 'base' is 100, we can only write 2 digits down in the answer section so the leading -1 of the -132 must be carried; thus:

${\displaystyle {\begin{matrix}111&&+11\\&&\downarrow \\88&&-12\\\hline &&-132\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}111&&+11\\&&\\88&_{-1}&-12\\\hline &&-32\end{matrix}}}$

Now we have -32 in the answer section. We must convert this negative number by replacing it with it's 'compliment', i.e. that number which when added would result in 100. In this case 32+68=100 so we replace the -32 with 68. Whenever we do a replacement of this sort we must also subtract one from the carry (i.e. in this case the carry changes from -1 to -2). Once this is done we continue as before; thus:

${\displaystyle {\begin{matrix}111&&+11\\&&\\88&_{-1}&-12\\\hline &&-32\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}111&&+11\\&&\\88&_{-2}&-12\\\hline &&68\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}111&&+11\\&\swarrow &\\88&_{-2}&-12\\\hline 99&&68\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}111&&+11\\&&\\88&&-12\\\hline 97&&68\end{matrix}}}$

111x88=9768
Note that in this example we have added the 11 to 88 during the Crosswise step because the 11 was written down as +11. It is important to note this sign. If we had instead used the other diagonal the calculation would have been 111-12, (resulting in the same answer 99), because the 12 is written as -12. Now all of this may seem a bit convoluted, but you would not explicitly write down each of the steps above, I have only done this to clearly illustrate what is going on. In practice the 'compliment and carry' steps would be written down directly, e.g.

• Multiply 97 x 104

${\displaystyle {\begin{matrix}97&\longrightarrow &-3\\\\104&\longrightarrow &+4\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}97&&-3\\&&\downarrow \\104&_{-1}&+4\\\hline &&88\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}97&&-3\\&\nwarrow &\\104&_{-1}&+4\\\hline 100&&88\end{matrix}}}$

97x104=10088
In the above case, after the initial differences are found, (-3 and +4), they are multiplied to give -12 but then you remember that a negative number cannot be written down so it is complemented giving 88 (12+88=100) and the carry is reduced by one, (since there is no carry in this example, the carry becomes -1). Finally the Crosswise step is performed, (in this case 97+4=101), and the carry subtracted (101-1=100), resulting in 100. The final answer being the concatenation of the two parts as usual, i.e. 10088.
Some further examples should make the process clear. Try doing them mentally before reading through the working.

• Multiply 103 x 87

${\displaystyle {\begin{matrix}103&\longrightarrow &+3\\\\87&\longrightarrow &-13\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}103&&+3\\&&\downarrow \\87&_{-1}&-13\\\hline &&61\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}103&&+3\\&\swarrow &\\87&_{-1}&-13\\\hline 89&&61\end{matrix}}}$

103x87=8961

• Multiply 998 x 1004

${\displaystyle {\begin{matrix}998&\longrightarrow &-2\\\\1004&\longrightarrow &+4\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}998&&-2\\&&\downarrow \\1004&_{-1}&+4\\\hline &&992\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}998&&-2\\&\nwarrow &\\1004&_{-1}&+4\\\hline 1001&&992\end{matrix}}}$

998x1004=1001992
In the above example the base is now 1000, the result of the vertically multiplication is -8 resulting in a compliment of 992.

• Multiply 1234 x 989

${\displaystyle {\begin{matrix}1234&\longrightarrow &+234\\\\989&\longrightarrow &-11\\\hline \quad \end{matrix}}\quad \Rightarrow \quad {\begin{matrix}1234&&+234\\&&\downarrow \\989&_{-3}&-11\\\hline &&426\end{matrix}}\quad \Rightarrow \quad {\begin{matrix}1234&&+234\\&\nwarrow &\\989&_{-3}&-11\\\hline 1220&&426\end{matrix}}}$

1234x989=1220426
The above example deserves some explanation as it shows the use of multiple techniques. First the initial residuals are written down, i.e. +234 and -11. We then use the special case technique to multiply 234 by -11 which results in -2574. This gives a carry of -2 and a remainder of -574. The remainder is complemented by subtracting it from 1000, this is done using the special case method for subtracting from a power of 10, (i.e. using the sutra All from 9 and the last from 10), resulting in 426 and the carry is reduced by one changing it from -2 to -3. Finally the Crosswise subtraction is performed taking account of the carry, (1234-11-3) resulting in 1220 giving a final answer of 1220426.

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