.62 for 2, D = 22 = 4
____
644 for 62, D = 2 x 6 x 2 = 24
32 for 62, D = 2(0 x 2) + 62
_____ = 36
3844
622 = 3844.Vedic Mathematics | Miscellaneous Items
MISCELLANEOUS ITEMS
1. Straight Squaring:
We have already noticed methods useful to find out squares of numbers. But the methods are useful under some situations and conditions only. Now we go to a more general formula.
The sutra Dwandwa-yoga (Duplex combination process) is used in two different meanings. They are i) by squaring ii) by cross-multiplying.
We use both the meanings of Dwandwa-yoga in the context of finding squares of numbers as follows:
We denote the Duplex of a number by the symbol D. We define for a single digit ‘a’, D =a2. and for a two digit number of the form ‘ab’, D =2( a x b ). If it is a 3 digit number like ‘abc’, D =2( a x c ) + b2.
For a 4 digit number ‘abcd’, D = 2( a x d ) + 2( b x c ) and so on. i.e. if the digit is single central digit, D represents ‘square’: and for the case of an even number of digits equidistant from the two ends D represent the double of the cross- product.
Consider the examples:
| Number | DuplexD | |
| 3 | 32 = 9 | |
| 6 | 62 = 36 | |
| 23 | 2 (2 x 3) = 12 | |
| 64 | 2 (6 x 4) = 48 | |
| 128 | 2 (1 x 8) + 22 = 16 + 4 = 20 | |
| 305 | 2 (3 x 5) + 02 = 30 + 0 = 30 | |
| 4231 | 2 (4 x 1) + 2 (2 x 3) = 8 + 12 = 20 | |
| 7346 | 2 (7 x 6) + 2 (3 x 4) = 84 + 24 = 108 |
Further observe that for a n- digit number, the square of the number contains 2n or 2n-1 digits. Thus in this process, we take extra dots to the left one less than the number of digits in the given numbers.
Examples:1 622 Since number of digits = 2, we take one extra dot to the
left. Thus
.62 for 2, D = 22 = 4
____
644 for 62, D = 2 x 6 x 2 = 24
32 for 62, D = 2(0 x
2) + 62
_____
= 36
3844
622 = 3844.
Examples:2 2342 Number of digits = 3. extradots =2 Thus
..234 for 4, D = 42 = 16
_____
42546 for 34, D = 2 x 3 x 4 = 24
1221 for 234, D = 2 x 2 x 4 + 32 = 25
_____
54756 for .234, D = 2.0.4 + 2.2.3 = 12
for ..234, D = 2.0.4 + 2.0.3 + 22 = 4
Examples:3 14262. Number of digits = 4, extra dots = 3
i.e
...1426
6, D = 36
________
1808246 26, D
= 2.2.6 = 24
22523 426, D =
2.4.6 + 22 = 52
_________
2033476 1426, D = 2.1.6 + 2.4.2
= 28
.1426, D = 2.0.6 + 2.1.2 + 42 = 20
..1426, D = 2.0.6 + 2.0.2 + 2.1.4 = 8
...1426, D = 12 = 1
Thus 14262 = 2033476.
With a little bit of practice the results can be obtained mentally as a single line answer.
Algebraic Proof:
Consider the first example 622
Now 622 = (6 x 10 + 2)2 =
(10a + b)2 where a = 6, b = 2
= 100a2 + 2.10a.b + b2
= a2 (100) + 2ab (10) + b2
i.e. b2 in the unit place, 2ab in the 10th place and a2 in the 100th place i.e. 22 = 4 in units place, 2.6.2 = 24 in the 10th place (4 in the 10th place and with carried over to 100th place). 62=36 in the 100th place and with carried over 2 the 100th place becomes 36+2=38.
Thus the answer 3844.
Find the squares of the numbers 54, 123, 2051, 3146.
Applying the Vedic sutra
Dwanda yoga.
2.CUBING
Take a two digit number say 14.
i) Find the ratio of the two digits i.e. 1:4
ii) Now write the cube of the first digit of the number i.e. 13
iii) Now write numbers in a row of 4 terms in such a way that the first one is the cube of the first digit and remaining three are obtained in a geometric progression with common ratio as the ratio of the original two digits (i.e. 1:4) i.e. the row is
1 4 16 64.
iv) Write twice the values of 2nd and 3rd terms under the terms respectively in second row.
i.e.,
1 4 16 64
8 32 (2 x 4 = 8, 2 x 16 = 32)
v) Add the numbers column wise and follow carry over process.
1 4 16 64 Since 16 + 32 + 6 (carryover) = 54
8 32 4 written and 5 (carryover) + 4 + 8 = 17
______________
2 7 4 4 7 written and 1 (carryover) + 1 = 2.
This 2744 is nothing but the cube of the number 14
Example 1: Find 183
Example 2: Find 333
Algebraic Proof:
Let a and b be two digits.
Consider the row a3 a2b
ab2 b3
the first is
a3 and the numbers are in the ratio
a:b
since a3:a2b=a2b:b3=a:b
Now twice of a2b, ab2 are 2a2b, 2ab2
a3 +
a2b + ab2 + b3
2a2b + 2ab2
________________________________
a3 +
3a2b + 3ab2 + b3 = (a + b)3.
Thus cubes of two digit numbers can be obtained very easily by using the vedic sutra ‘anurupyena’. Now cubing can be done by using the vedic sutra ‘Yavadunam’.
Example 3: Consider 1063.
i) The base is 100 and excess is 6. In this context we double the excess and then add.
i.e. 106 + 12 = 118. (
2 X 6 =12 )
This becomes the left - hand - most portion of the cube.
i.e. 1063 = 118 / - - - -
ii) Multiply the new excess by the initial excess
i.e. 18 x 6 = 108 (excess of
118 is 18)
Now this forms the middle portion of the product of course 1 is carried over, 08
in the middle.
i.e. 1063 = 118 / 08 / - - - - -
1
iii) The last portion of the product is cube of the initial excess.
i.e. 63 = 216.
16 in the last portion and 2 carried over.
i.e. 1063 = 118 / 081 /
16 = 1191016
1 2
Example 4: Find 10023.
i) Base = 1000. Excess = 2. Left-hand-most portion of the cube becomes 1002+(2x2)=1006.
ii) New excess x initial excess = 6 x 2 = 12.
Thus 012 forms the middle portion of the cube.
iii) Cube of initial excess = 23 = 8.
So the last portion is 008.
Thus 10023 = 1006 / 012 / 008 = 1006012008.
Example 5: Find 943.
i) Base = 100, deficit = -6. Left-hand-most portion of the cube becomes 94+(2x-6)=94-12=82.
ii) New deficit x initial deficit = -(100-82)x(-6)=-18x-6=108
Thus middle potion of the cube = 08 and 1 is carried over.
iii) Cube of initial deficit = (-6)3 = -216
__
__
Now 943 =
82 / 08 / 16 = 83 / 06 / 16
_
1 2
= 83 / 05 / (100 – 16)
= 830584.
Find the cubes of the following numbers using Vedic sutras.
103, 112, 91, 89, 998, 9992, 1014.
3. Equation of Straight line passing through two given points:
To find the equation of straight line passing through the points (x1, y1) and (x2, y2) , we generally consider one of the following methods.
1. General equation y = mx + c.
It is passing through (x1, y1) then y1 = mx1 + c.
It is passing through (x2, y2) also, then y2 = mx2 + c.
Solving these two simultaneous equations, we get ‘m’ and ‘c’ and so the equation.
2. The formula
(y2 - y1)
y – y1 = ________ (x – x1) and substitution.
(x2
- x1)
Some sequence of steps gives the equation. But the paravartya sutra enables us to arrive at the conclusion in a more easy way and convenient to work mentally.
Example1: Find the equation of the line passing through the points (9,7) and (5,2).
Step1: Put the difference of the y - coordinates as the x - coefficient and vice - versa.
i.e. x coefficient = 7 - 2 = 5
y coefficient = 9 - 5 = 4.
Thus L.H.S of equation is 5x - 4y.
Step 2: The constant term (R.H.S) is obtained by substituting the co-ordinates of either of the given points in
L.H.S (obtained through step-1)
i.e. R.H.S of the equation is
5(9) -
4(7) = 45 - 28 = 17
or 5(5) - 4(2) = 25 - 8 = 17.
Thus the equation is 5x - 4y = 17.
Example 2: Find the equation of the line passing through (2, -3) and (4,-7).
Step 1 : x[-3-(-7)] –y[2-4] = 4x + 2y.
Step 2 : 4(2) + 2(-3) = 8 –6 = 2.
Step 3 : Equation is 4x + 2y =2 or 2x +y = 1.
Example 3 : Equation of the line passing through the points (7,9) and (3,-7).
Step 1 : x[9 - (-7)] – y(7 - 3) = 16x - 4y.
Step 2 : 16(7) - 4(9) = 112 – 36 = 76
Step 3 : 16x- 4y = 76 or 4x – y = 19
Find the equation of the line passing through the points using Vedic methods.
1.
(1, 2), (4,-3)
2. (5,-2), (5,-4)
3. (-5, -7), (13,2)
4. (a, o) , (o,b)