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SECTIONAL TIPS
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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
2. Write down the answer one figure
at a time, from right to left as
in
any
multiplication.
The figures of the answer are obtained
by adding to each
successive
digit of the multiplicand. its
right neighbour. Remember the
right
neighbour is the right, (i.e., the
correct) neighbour to be
added.
(1) 123 X 11 = ?
Step1: Prefix a zero to the
multiplicand so that it reads 0123.
Step2: To the right digit 3,
add its right
neighbour.
There is no neighbour on the right; so add 0. 0123 X 11
3 + 0
=3.
3
3
To the next digit 2, add
its right neighbour 3.
0123 X 11
2 + 3 = 5.
53
To the next digit 1, add the right neighbour 2. 0123 X
11
1 + 2=3.
353
To 0, add the right neighbour 1.
0123 X
11
0 + 1 = 1.
1353
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.
Step2: Double 3 and add the right
neighbour 013 X
12
neighbour;therefore add0).
6
Double 1 and add the right neighbour 3.
013 X 12
2 X 1 + 3 =5 56
Double
0 and add the right neighbour 1.
013
X 12
2 X 0 + 1 =1
156
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 =
?
2331
21132
039942 X
13
043285 X
14
519246
605990
(3)
58265 X 15 = ?
(4) 36987 X 16 =
?
34132 25654
058265 X 15 036987
X 16
873975
591792
(5)
69873 X 17 = ?
(6) 96325 X 18 =
?
57652 85224
069873 X 17
096325 X 18
1187841 1733850
(7)
74125 X 19 = ?
73124
074125 X 19
1408375
2. Multiplication of two 2 digit numbers
Consider the conventional
multiplication of two 2 digit numbers 12 and
23
shown
below:
12
X 23
36
24
Ans.
276
It is obvious from the
above that
(1) the
right digit 6 of the answer is the product obtained by the
"vertical"
multiplication of
the right digit of the multiplicand
and of the multiplier.
(2) the left digit 2 of the
answer is the product obtained by the "vertical"
multiplication of the left digit of the multiplicand and of
the multiplier
(3) the middle digit 7 of the answer is
the sum of 3 and 4. The 3 is the product
of the left digit of the multiplicand and the
right digit of the multiplier; the
4
is the product of the right
digit of the multiplicand and
the left digit of the
multiplier.
This means that, to obtain the middle digit, one has to multiply
"across" and add the two
products (in our example 1 X 3 + 2 X 2)
The working in
our above example can therefore be depicted as
1
2
2 3
1 X 2 / 1 X 3 + 2 X 2
/ 2 X 3
1 2
2
3
2 / 3 + 4 / 6 = 276
3.
When the units figure is "one" in both the numbers being
multiplied, the
process of multiplication is simplified
further. Consider the following
multiplication:
31
21
2 X 3 / 2 X 1 + 1 X 3
/ 1 X 1
You will notice that the middle digit of the answer is 2 X 1 + 1 X 3
i.e.
(2 + 3) X 1. So, instead of
multiplying "across" for the middle term,
you
could simply add the tens
digit of the two numbers.
Therefore, 31 X 21 = 6 / (2 + 3) / 1 =
651
Similarly, in 81 X 91, you could obtain the middle term as
17, by
merely
adding 8 and 9.
4. Like wise, when the tens figure is
"one" in both the numbers being multiplied,
you could obtain the middle term by
simply adding the units digit of the two
numbers. For instance, the middle term in 12 X 17 is 2 +
7, i.e. 9, in 8 X 12
it is 8 + 2 i.e. 10
etc.
5. If the units
figure or the tens figure is the same in the two numbers,
the
process of multiplication could be
simplified as shown in the following
examples:
(1)
8 3 The middle term is obtained by
multiplying 3 by 17, 17 being he sum of 8 9 and
93
8
3
9
3 and
93
72519 =
7719
(2) 2 8 In this case, the middle term is 2
multiplied by 11, 11 being the sum of 8 2 3 and
3
42 22
4 =
644
6. Multiplication of 2
three-digit numbers
Let us consider the multiplicand to be ABC and the multiplier
to be DEF, as
shown below:
A B C
X D E
F
Answer
1. The extreme right digit
of the answer is obtained (as before) by
vertical multiplication as
C X F
2. The extreme left digit
is also obtained (as before) by vertical
multiplication as A X B
3. The "middle" digits are
obtained (as before) by multiplying
across.
Progressing one step at a time to
the left, the "middle" digits
are
successively
B X F + C X E
A X F + C X D + B X E
A X E
+ B X D
The process is set out in detail for following examples below
.
1.
1
2
3
4
5
6
1 X 4 / 1 X 5 + 2 X 4 / 1 X 6 + 3 X 4 + 2 X 5 / 2 X 6 + 3 X 5 / 3 X
6
= 4/13/ 28/ 27 / 18 =
56088
2.
2
4
5
1
9
8
2 X 1/ 2 X 9 + 4 X 1 / 2 X 8 + 5 X 1 + 4 X 9 / 4 X 8 + 5 X 9 / 5
X 8
= 2/ 22/
57 / 77/ 40 = 48510
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.
The
following examples will clarify the procedure:
(1)
286 X 78 =?
(2)
998 X 98 =?
2 8 6
9 9 8
0 7 8
0 9 8
0
1472 106 48 = 22308 0
81 153 144 64 = 97804
8. Special
case where the units figures of the multiplicand and the multiplier
together total 10 and the other figures are the
same.
Consider
(i)
23 X 27
(ii)
94 X 96
(iii)
982 X 988
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
extreme right)
digit of the 2 numbers
(2) To obtain the left
part of the answer, multiply the other (i.e. the
left),
digit/s by one
more than itself/themselves.
For example, if the left
digit/s
of the 2
numbers is/are
3, multiply 3 by (3 + 1) i.e. by
4
5, multiply 5 by 6
6, multiply 6 by 7
99,
multiply 99 by
100
100,
multiply 100 by
101
888,
multiply 888 by 889
etc
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
numbers to be
multiplied
b. Put the two
numbers A and B above and below on the left hand
side
c. Subtract each of
them from the base (nearest power of 10) and
write
down the remainders
r1and r2 on the right hand side either with
a
connecting minus sign between
A & r1 and B & r2 if the
numbers A and B
are
less than the closest power of 10. Otherwise, use a connecting
plus
sign between the numbers and
the remainders.
d. The final answer will
have two parts. One on the left hand side and
the
other on the right hand side. The right hand side is the
multiplication of
the two
remainders and the left hand side is
either the difference of A
and
r2 or B and
r1 if the numbers are less than the closest power of 10.
Otherwise, it is the sum of A and r2 or B and
r1
e.q. 1. Multiplication of 9 and
7
The closest base to the two numbers in this case is
10
Therefore 9 - 1
(The remainder after subtracting the number from 10)
7 – 3 (The remainder
after subtracting the number from 10)
The right hand side of the answer
will be 1 X 3 = 3
The left hand side can be computed either by
subtracting 3 from 9 or 1 from 7 which is 6. Therefore, the answer is
63.
e.q. 2. A more difficult example will be the
product of 94 and 87
The closest base in this case will be 100
Therefore
94 – 6
87 – 13
Here, 6 in the first row is the difference 100 and 94 and the 13
in the second row is the difference between 100 and 87. The right hand
side of the answer is obtained by the multiplication of 6 and 13 which is
78 and the left hand side is obtained by the difference between either 87
and 6 or 94 and 13, both of which give the answer 81.
e.q. 3. Taking numbers which are greater than the closest power of 10
Find the product of 108
and 112
The closest base is 100 in this case as
well
Therefore
108 + 8
112 + 12
e.q. 4. When the number of digits of the product of
the remainders is greater than the power of
e.g. Product
84 and 92
84 – 16
92 – 8
As the product of 16 and 8 is 128 which is a three digit number as
against 2 being the power of 10 in100, we carry forward the digits on the
left more than 2 digits (in this case) and add to 76, the left hand side
of the answer
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
Product of 88 and 106
88 – 12
108 + 8
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
Multiplication of numbers which are not close to the nearest power
of 10
Method 1: Take 50 as the base
which is half of 100
41 – 9
43 –
7
34 .63 à
Since 50 = 100/2. we divide the left hand side number also by 2 while
retaining the right hand side. Therefore the answer will be
1763.
Method 2
43 + 3
To Calculate the squares from 51 to
59:- There is one general formula
(5x)2 = (25 +
x)/x2
Where x is a digit in units place
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
Let us calculate the squares of 54
In this case
our x is 4
(54)2 = (25 + 4)/42 = 29/16 = 2916 is
our answer
TO
CALCULATE THE PRODUCT OF NOs. (Ending with 5) WHICH DIFFER BY 10
e.g. 45 x 35
Step
1: First of all
write 75 in the last two places of the product 45
x
35 =
__75
Step2:
Then Multiply 4 with 3 and add the smallest of these to the
product
i.e. (4 x 3 + 3) = 15
Step3: Now Place these two digit
before 75 to get our required product
i.e.
45 x 35 = 1575