CRISS-CROSS ONE-LINE FASTEST MULTIPLICATION TRICK | multiply any no. mentally in 10-15 sec

Sohel Sahoo: Hello guys,

In this article we will learn an universal formula for multiplication,which is conventional but described in unconventional way here after a long process of research.Urdhva tiryagbham, is the general formula applicable to all cases.

One must note that the two no.s which are being multiplied together should have equal digits;if not then make the lesser one i.e. multiplier equal to multiplicand by adding required no of zeros before it.In this way this is universal &takes least time. 

(a) Let the two 2 digit numbers be (ax+b) and (cx+d), Note that x = 10. 

Now,
Consider the product of two numbers
(ax + b) (cx + d)
= ac.x² + adx + bcx + b.d

 = ac.x² + (ad + bc)x + b.d 

Observe that:-

(i) The first term i.e., the coefficient of x² (i.e., 100, hence the digit in

the 100th place) is obtained by vertical multiplication of a and c i.e., the
digits in 10th place (coefficient of x) of both the numbers;

(ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10th place)
is obtained by cross wise multiplication of a and d; and of b and c; and the addition of the two products;

(iii) The last (independent of x) term is obtained by vertical
multiplication of the independent terms b & d.

(b) 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)

 x (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

Note the following points :

 (i) The coefficient of x⁴ , i.e., ad is obtained by the vertical

multiplication of the first coefficient from the left side :

(ii)The coefficient of x³ , i.e., (ae + bd) is obtained by the cross –wise

multiplication of the first two coefficients and by the addition of the two
products;

(iii) The coefficient of x² is obtained by the multiplication of the first coefficient of the multiplicand (ax2+bx +c) i.e., a; by the last coefficient

of the multiplier (dx2 +ex +f) i.e.,f ; of the middle one i.e., b of the
multiplicand by the middle one i.e., e of the multiplier and of the last
one i.e., c of the multiplicand by the first one i.e., d of the multiplier
and by the addition of all the three products i.e., af + be +cd :

(iv) The coefficient of x is obtained by the cross wise multiplication of the
second coefficient i.e., b of the multiplicand by the third one i.e., f of
the multiplier, and conversely the third coefficient i.e., c of the
multiplicand by the second coefficient i.e., e of the multiplier and by
addition of the two products, i.e., bf + ce ;

v) And finally the last (independent of x) term is obtained by the
vertical multiplication of the last coefficients c and f i.e., cf

Thus the process can be put symbolically as (from left to right)

Consider the following example:-124 X 132.

Proceeding from right to left
i) 4 X 2 = 8. First digit = 8

ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is
carried over to left side. Second digit = 6.

iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of
above step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is
carried over to left side. Thus third digit = 3.

iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is
added i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6

v) ( 1 X 1 ) = 1. As there is no carried over number from the previous
step it is retained. Thus fifth digit = 1
Henceforth,124 X 132 = 16368.

To understand the characteristics of this method,carefully observe the steps used in multiplying 2-digit,3-digit,4-digit&5-digit numbers as given below.

                                                                                  

Technique for multiplying 6-digit no.s

Likewise we can construct the rest multiplication of higher numbers .

Note :

1. We can carry out the multiplication in urdhva - tiryak process from left to right or right to left.

2. The same process can be applied even for numbers having more digits, provided the no. having lesser digit should be made equal with other by adding required no.of zeros before itself.

• 3. urdhva –tiryak process of multiplication can be effectively used in multiplication regarding algebraic expressions & when a power of x is absent,it should be given a zero coefficient.

  Find (3x²+ 4x + 7) (5x +6)

  Now, 3x²+ 4x + 7
     x  0.x²+ 5x + 6
 ¯¯¯¯¯¯¯¯¯¯¯¯
i) 7 X 6 = 42
ii) (4 X 6) + (7 X 5) = 24 + 35 = 59 i.e., 59x
iii) (3 X 6) + (4 X 5) + (7 X 0) = 18 + 20 + 0 = 38 i.e., 38x²
iv) (3 X 5) + (0 X 4) = 15 + 0 = 15 i.e., 15x³
v) 3 X 0 = 0
Hence the product is 15x³ + 38x² + 59x + 42

Hope this will serve my purpose of writing to reveal the logic &concepts behind each word of mathematics&at the end you shall be affectionated with it.
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