Vertically and Cross wise

Vertically and cross wise Sutra : It is also called as  Urdhva – tiryagbhyam  which is the general formula applicable to all cases of multiplication and also in the division of a large number by another large number.

(a) Multiplication of two 2 small digit numbers.

Lets take two  numbers,say 15 x 23

Here start the multiplication from Right to Left

Step :1 

1   5

      |

2   3

------

 5 x 3 =15 ......keep the last digit  '5' and carry over '1'

Step 2:

1  5

  \/

  /\

2  3

-------

(1 x 3 ) + (2x5) = 3+10 =13

With carry over '1' from step 1 ,we get 13+1 =14.....keep the last digit '4' and carry over '1'

Step 3:

1  5

|

2  3

--------

1 x 2=2

With carry over '1' from step 2, we get 2+1=3

So our result from step 1 to step 3 = 345

Therefore, 15 x 23 =345

(b) Multiplication of two 2  large digit numbers

 With the number 72 x 88 ,We do all the above three steps from (a)

   7  2

   | \/ |

   | /\ |

   8  8

-------------

7x8 / ( (7x8)+(2x8)) / 2x8

56 / ( 56 +16) /16

56 / 72 / 16

 =  6336   , by keeping the last digits and carrying over the rest

Therefore, 72 x 88 = 6336

Algebric Proof :

(ax + b) (cx + d)  =  ac.x² + adx + bcx + b.d

                            =  ac.x² + (ad + bc)x + b.d

where in our example  15 x 23 , a=1 ,b=5 ,c=2,d=3

and so ac.x² + (ad + bc)x + b.d = 2 x²  +13 x +15  where x=10 meaning the place value

                                                 =   345 
 by  keeping the last digits and carrying over the rest.

Now consider the multiplication of two 3 digit numbers.

Let the two numbers be (ax² + bx + c) and (dx² + ex + f). Note that x=10
Now the product is
                            ax² + bx + c    multiply
                            dx² + ex + f

          ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
         ad.x⁴+bd.x³+cd.x²+ae.x³+be.x²+ce.x+af.x²+bf.x+cf
=       ad.x⁴ + (bd + ae). x³ + (cd + be + af).x² + (ce + bf)x + cf

From the above expression,  we see that we multiply from left to right as per the expression

Example :
Find the result of  234 x 615 ?

ad.x⁴ + (bd + ae). x³ + (cd + be + af).x² + (ce + bf)x + cf  where  a= 2,b=3,c=4 and d=6 ,e=1,f=5

=2x6 +( 3x6 +2x1 ) +(4x6+3x1 +2x5) +(4x1+3x5)+ 4x5...lets keep X's outside
=12 +(18+2) +(24+3+10)+(4+15)+20
=12x⁴+20x³+ 37x² +19x+20  ..inserting X's now

And so keeping the last digits and carrying over the rest, we get   143910

Therefore,  234 x 615 =143910

Nice ,as the proverb "Practicing keeps a Man perfect " ,practicing the above sutras makes your multiplication easier .Keep practicing..that's the Veda of Vedic Maths

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