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Vedic Math | |||
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1. Special method for multiplication by numbers from 11 to 19. Multiplication by 11 Rule: 1. Prefix a zero to the
multiplicand
To the next digit 2, add
its right neighbour 3.
0123 X 11 Therefore, 123 X 11 = 1353 (which you can easily verify by a conventional multiplication). MULTIPLICATION BY 12: The method is exactly the same as in the case of 11 except that you double each number before adding the right neighbour. (1) 13 X 12 =? Step1: Prefix a zero to the
multiplicand so that it reads 013.
Double 1 and add the right neighbour 3.
013 X 12 Double
0 and add the right neighbour 1.
013
X 12 Therefore, 13 X 12 = 156 (which you can again verify by a conventional multiplication).
MULTIPLICATION FROM 13 TO 19: The reason why the rule is different for multiplication by 11 and by 12 is obviously because the right digits are different. The right digit, we could call the Parent Index Number (PIN). Thus in 11, the PIN is 1 and in 12 it is 2. (In 13, it is 3; in 14, it is 4 etc.) When the PIN is 1, we are simply taking each figure of the multiplicand (we could call this figure the Parent Figure – PF in short ) as such and adding the right neighbour. When the PIN is 2, we are doubling the PF and then adding the right neighbour. Obviously, if the PIN is
3 (as in 3), we would treble the PF and then add the right neighbour. If
the PIN is 4 (as in 14), we would quadruple (i.e. multiply by 4) the PF
and then add the right neighbour. If PIN is 9 (as in 19), we would
multiply the PF by 9 and then add the right neighbour, What is the
advantage of the method? We need know the tables only upto 9 and still
multiply by a simple process of addition. (1) 39942 X 13 =
?
(2) 43285 X 14 =
? (3)
58265 X 15 = ?
(4) 36987 X 16 =
? (5)
69873 X 17 = ?
(6) 96325 X 18 =
? (7)
74125 X 19 = ? 2. Multiplication of two 2 digit numbers Consider the conventional
multiplication of two 2 digit numbers 12 and
23 It is obvious from the
above that (2) the left digit 2 of the
answer is the product obtained by the "vertical"
(3) the middle digit 7 of the answer is
the sum of 3 and 4. The 3 is the product The working in
our above example can therefore be depicted as
1 2 3.
When the units figure is "one" in both the numbers being
multiplied, the 31
You will notice that the middle digit of the answer is 2 X 1 + 1 X 3
i.e.
(2) 2 8 In this case, the middle term is 2
multiplied by 11, 11 being the sum of 8 2 3 and
3 6. Multiplication of 2
three-digit numbers
Let us consider the multiplicand to be ABC and the multiplier
to be DEF, as 1. The extreme right digit
of the answer is obtained (as before) by
2. The extreme left digit
is also obtained (as before) by vertical
3. The "middle" digits are
obtained (as before) by multiplying
across. 2.
2
4
5 7. Multiplication of Numbers of Different Lengths In the examples we saw
above, both the multiplicand and the multiplier contained the same number of digits. But what if the
two numbers were to contain a different number of digits; for instance, how would we multiply 286
and 78? Obviously we could prefix a zero to 78 (so that it becomes 078, a 3 digit number)
and proceed as in any multiplication of two 3 digit numbers. (1)
286 X 78 =?
(2)
998 X 98 =? 8. Special
case where the units figures of the multiplicand and the multiplier
together total 10 and the other figures are the
same. In all of them, the units figures together total 10 [3 + 7] in (i), 4 + 6 in (ii) and 2 + 8 in (iii)], also the other figures of the multiplicand and the multiplier are the same [2 in (i) 9 in (ii) and 98 in (iii)] In such cases, a special method can be used. But before we see the new method, let us consider the multiplication of 23 by 27 by the method we had learnt. 23 X 27 would give us the answer as 621. Do you notice anything special about the answer? The right part is the product of the unit figures 3 and 7 of the 2 numbers and the left part 6 of the answer is the product of 2 (the tens figure) and the next higher number 3. This gives us the special rule: (1) To obtain the right
part of the answer, multiply the units (i.e. of
the (2) To obtain the left
part of the answer, multiply the other (i.e. the
left), 3, multiply 3 by (3 + 1) i.e. by
4 The only thing you have to be careful about is to ensure that the right part of the answer always has 2 digits. For example in 29 X 21, the left part of the answer is 2 (2 + 1) and the right part is 9 X 1 i.e. 9; however the right part will be written as 09 and not as 9 because the right part has to contain 2 digits. Therefore, 29 X 21 = 609. 9. Special Method for Squaring numbers ending in 5 An obvious extension of the above method will be in finding the square of numbers ending in 5. Consider 15 X 15. Here the sum of the units digit of the 2 numbers is the same. Hence, the above method will apply. Infact it will apply in squaring any number ending in 5. Since in all these cases the right digit of the multiplicand and the multiplier is 5, the right part of the answer is always 25 and therefore, we can mechanically set down 25 as the right part of the answer without any calculation – all that is needed is to find out the left part of the answer and this done exactly as in the previous section. 152 =
225
252 = 625
8752 = 765625 89952 =
80910025 TO MULTIPLY ANY TWO
NUMBERS A AND B CLOSE TO A POWER OF 10
a. Take as base for the
calculations that power of 10 which is nearest to
the e.q. 1. Multiplication of 9 and
7 e.q. 3. Taking numbers which are greater than the closest power of 10 Find the product of 108
and 112 e.q. 5, When one of the
number is lesser than the closest power of 10 and the other greater than
the closest power of 10 The operation is similar,
excepting that as the right hand side of the answer is obtained by the
multiplication of a positive
and a negative number the answer has to be subtracted from 100 by reducing
the left hand side number by 1 Method 1: Take 50 as the base
which is half of 100 Method 2 To Calculate the squares from 51 to
59:- There is one general formula Note: Slash here does not
represent division, It is just to differentiate
between the two value. x2 should be a two digit no. If it is a
single digit no, make it a two digit no by placing a prefix 0 before
x2 TO
CALCULATE THE PRODUCT OF NOs. (Ending with 5) WHICH DIFFER BY 10
e.g. 45 x 35
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