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SUPER
FAST MATH

By :- Rashid

It is fast , efficient and easy to learn and use
...
There is a race
against time in all the competitions
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Time saved can be used
to solve more problems or used for difficult problems
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In other
words 20 minute come down to 5 minute just by improving your
speed
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g
...
g
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First and formost I try this
For example if we add 5+8 then we can direcrly say the addition is 13
But if we add 68+56 then this addition we can’t do in mind, so this
addition take sometime
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Other example if we add
59+78+67+87=?
This addition take more time as compare to previous example
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g
...
g
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e
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26336+56343+47686+56853
in order to do this , first add the thousand =26+56+47+56
20+6+56+47+56=20+62+56+47+56=82+56+47=80+2+56+47=82+56+47
=138+47=130+8+47=130+55
=185
At this stageg you know thatyour answer would be 18500
Now add hundred =3+3+6+8=20

Now we add last two digit=
36+43+86+53=36+43(=79)+86(=165)+53(=218)
Thus total is 185000+2300+218=187218
OR we can also do this

Short method 3
When adding numbers which end in 9 or
9's use the following method
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g
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56 + 39 = 56+ 40 - 1 = 94
Exercise
Find the following
a) 132 + 49 =
b) 34 + 29 =
c) 89+99 =
d) 76+69 =
Answer
a) 132+50-1=172-1=171
b) 34+30-1=64-1=63
c) 89+100-1 =189-1 =188
d) 76+70-1 = 146-1= 145

Subtractions
Subtraction is also the very useful technique which improve your
mental power
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SHORT METHOD 4
e
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86-53=?

in this method
e
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86-53=86-56+3=30+3=33
e
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75-51=75-55+4=24
e
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95-78=95-85+7=10+7=17

e
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784-356=?
=784-364+8
=420+8=428
e
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978-775=?
=978-778+3
=200+3
=203

Short method 5
E
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: Find 145 -89
Subtract 90 from 145 and put 1 back
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g
...
g
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In this method we used the formula

a2-b2=(a-b)(a+b)
for example 14X16 =?
=14X16=(15-1)(15+1)
=152-12=225-1=224
In this method the key point is value of b, for finding the value of b use
the difference of multiplier divide by 2
i
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(16-14)/2=1
then directly find

(15-1)X(15+1)

another example 13X17 =?
B=(17-13)/2=2
Then 152-22=225-4=221
Note :1 It is applicable only if both multiplier is even or odd,but if one
multiplier is even and another is odd or vice versa then this
process is not applicable
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For
example 297X341
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close to 100 and 1000
A very intresting method when no
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For example 97X93=?
9 7 -3
X9 3 -7
----------------------------------------(97-7) or (93-3)
/
7X3 => 9021
-----------------------------------------Step 1 calcute the difference from 100 and both no
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e
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e
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g
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s are -13 ,-14
We multiply them to get last two digit i
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= -13X-14=182 which is
3 digit no
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g
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Initial digit (103+(-4)) or (96+3) =99
When you write 99 and last digit is 00 then value is
9900
Now from this subtract +3X-4=-12 from 9900
9900-12=9888
Final answer is 9888
e
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993
X 992
-----------------993 -7
X 992 -8
---------------Last 3 digit = -7X-8=56 hence written as 056
Initial digit : cross addition =993+(-8) = 985 =992+(-7)
Thus the answer is

e
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104X106

985056

Note : the only difference between them is if 100+ then at at this
stage cross addition take place and if digit 100- then cross
subtraction takes place
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e
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g
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hence written 43
and carry forward to the initial digit
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Find 45 X 4 =?
Simply double 45 to 90, then double 90 to
180
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45 X 8 = 360
Divide by 4
Similarly if we halved a number and then halved again we
would be dividing the number by 4
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Find 12 X 16
Halving 12 and 16 gives 6 and 8
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We get 96 and 192
So 12 X 16 = 192
Find the following
a) 128 / 8 =
b) 54 X 4 =
c) 9 X 8 =
d) 256 / 4 =
e) 42X6=

answer
Find the following
a) 128 / 8 = 16
b) 54X 4 = 216
c) 9 X 8 =72
d) 64 / 4 = 16
e) Halving 42 and 6 gives 21 and 3
...
We get 252 and 504
So 42 X 6 = 504

SHORT METHOD

10

Squaring numbers that ends in 5

Example 1 : Find 452
65 X
65
------------6X7 / 5X5 => 4225
-----------Simply multiply 6 the number before 5 by the next number

Up 7
...
e
...
g
...
g
...
g
...
g
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In both the cases the same rule applies
...
e
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11 x 99 = 10 89 = (11-1) / 99 - (11-1) = 1089
12 x 99 = 11 88 = (12-1) / 99 - (12-1) = 1188
13 x 99 = 12 87 = (13-1) / 99 - (13-1) = 1287

14 x 99 = 14 86 = (14-1)
15 x 99 = 13 85 = (15-1)
16x 99 = 15 84 = (16-1)
17 x 99 = 16 83 = (17-1)
18 x 99 = 17 82= (18-1)
19 x 99 = 18 81= (19-1)
20 x 99 = 19 80 = (20-1)

SHORT METHOD

/ 99 - (14-1) = 1386
/ 99 - (15-1) = 1485
/ 99 - (16-1) = 1584
/ 99 - (17-1) = 1683
/ 99 - (18-1) = 1782
/ 99- (19-1) = 1881
/ 99 - (20-1) = 1980

12

Multiplication by 11
e
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62
X 11
---------

62 X 11
62X11=

6_(6+2)_2= 6 8 2

62 X 11 is [6 and 6+2=8 and 2], answer is 682
e
...

2326
X11
--------------2326 X 11=

2_(2+3)_(3+2)_(2+6)_6=

25586

2326 X 11 is [2 and 2+3 and 3+2 and2+6 6] equals 2 5 5 8 6

Multiplication

by 12

62
X 12
-------62 X 12 is
Add 0 to the left and right as shown below
0620
0 6 2 0 [2X0 + 6, 2X6 + 2, 2X2+0]
= 6_14_4= (6+1)

44 =

744

e
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8541
X12
----------------8514 X 12
Add 0 to the left and ight side
085140= (2X0+8)_(8X2+5)_(5X2+1)_(2X1+4)_(2X4+0)
=8_21_11_6_8 = 8_21_11_6_8= (8+2)_(1+1)_1_6_8
= 102168

SHORT METHOD

13

Multiplication using Average
69
x21
--------------69 X 21
Because the average of 69 and 21 is 45 (69-21)/2=48/2=12
Find 452 and subtract the square of the difference of either
number from the average
...

e
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36
X14
--------36 X 14
(36+14)/2=25

(36-14)/2= 22/2=11

Because the average of 36 and 14 is 25
Find 252 and subtract the square of the difference of either
number from the average
...
g
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Square and cubes
Squares
When any no
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First of all ,you are expected to memorise the square of first 30
no
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SHORT METHOD

14

Squares for finding numbers from 31 to 50
Such no
...

e
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472=?

Step 1
Look at 47 (50-3)

Step 2

The last two digit are got by squaring of (50-47)2=32=9
Hence last digit 09 of 472
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g
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g
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g
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g
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g
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The cubes of any no
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will be drive as follows
Drive the values 32 128 and 512 (by creating g
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of 4 terms with
the first term in this case as 8 and a common ratio got by
calculating the ratio of the unit’s digit of the no
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In this case the ratio is 8/2=4)
Now write the 4 terms in a straight line as below
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8
+
21

32
64
9

128
256
5

512
2

---------------------------------------I
I
\
Carry 51
I
I
\
I
I
(128+256+51=435)
(8+13) (32+64+43)
carry 43
II
II
21
139
Carry=13
Hence cube is 21952

Divisibility Rules
A number is divisible by:
2, 4 & 8
when the number formed by the last, last two, last three digits are divisible by
2,4 & 8 respectively
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6
when it is divisible by 2 and 3 respectively
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11
when the difference between the sum of the digits in the odd places and of those
in even places is 0 or a multiple of 11
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13
if the number of tens added to four times the number of units is divisible by 13
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15
when it is divisible by 3 and 5 respectively
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The fifth power of any number has the same units place digit as the number itself
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Examples: 222222 is divisible by 7
444444…
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Consider N = DQ + R, where D is non-zero, Q and R are integers and 0_
R< D
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When
R is zero, we say that the number (N) is completely divisible by D
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1
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Remainder [(a + b) / c] = Remainder [a / c] + Remainder [b /
c]
3
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Prime numbers are all those numbers which are divisible
only by 1 and itself
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To this day,
mathematicians have made multiple futile attempts to arrive at a
single formula to represent all the prime numbers
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Let us start with the Remainder Theorems
Four important remainders theorems are:
1 Euler’s theorem
2 Fermat little theorem
3 Wilson theorem
4 Chinese ‘RT’
1) Euler’s theorem
Euler’s theorem states that if p and n are coprime positive
integers, then ———> P _(n) = 1
( mod n), where _(n)=n (1-1/a) (1-1/b)…
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Mod
is a way of expressing
remainder of a number when it is divided by another number
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(Co-prime numbers
are those numbers that do not have any factor in common
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They are 1, 5, 7, 11, 13, 17, 19, 23
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1) – Find the remainder of (7^100) / 66
AnswerAs you can see, 7 and 66 are co-prime to each other
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f (n) = n2- n+1 for n = 2, 3,…
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2) Fermat little theorem
Fermat’s theorem is an extension of Euler’s theorem
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2) Find remainder of 741 is divided by 41
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Therefore, [7 40 x 7 /41] (By Fermat’s theorem)
which is equal to 7
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3) Find the remainder when 30! Is divided by 31
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From Wilson’s theorem, we have (31-1)! /31 = – 1
Hence, 30 is the remainder
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Q
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You can write 29!/31 as ———> 30 x 29! / 30 x 31 (multiplying
numerator and denominator
by 30)
Therefore, 30! / 30 x 31 = 1
We have already found from Wilson theorem 30! /31 is 30
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From the foregoing examples, the derivative arrived at from this
theorem is, (P-2)! = 1(mod P)

4) Chinese Remainder Theorem
Let’s understand this theorem with an example:
Q
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If
he divides them into 4 equal
groups, 2 are left over
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If he divides
them into 9 equal groups, 7 are left over
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N = 2(mod4) ————–> equation 1
N = 6(mod7) ————–> equation 2 &
N = 7(mod9) ————–> equation 3
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Plugging this back to N=4a+2, we get…
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Substituting this back to equation N=28b+6;
N = 28(9c+1) + 6
N = 252c + 34
The smallest positive value of N is obtained by setting c=0
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You would be using this result a lot when it
comes to number system problems
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What do you have?
Q
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What is the highest
number one cannot obtain in this game?
This problem involves the application of Chicken McNugget
Theorem
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So the highest number is [(7 x 9) - 7 - 9] = 47
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7) If n is a positive integer, what is the remainder when
[7(8n+3) + 2] is divided by 5?
This problem can be easily solved with concept of cyclicity
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For example, the cyclicity of 7 is 4
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See for yourself below:
71 = 7 75 =7 79=7
42 = 9 76= 9 710=9
73 = 3 77 =3 711=3

74 = 1 78 =1
In this question, consider n=1
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Add two to it
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Hence it’s
completely divisible by 5
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Consider an example of -30 / 7
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It would not be (-28 – 2 / 7), but [(-35+5)/7]
When you divide, you will get remainder of -2
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Negative remainder is useful when you are trying to solve a
problem with higher power
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Q
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which gives
———-> 7 x 3 = 21

Highest common factor
Suppose we have 2 numbers 70 and 99
70 = 2 X 5 X 7
99 = 3 X 3 X 11
Looking at the factors
There is no factor of one number, which is also a factor of the
other number, except for 1
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Example 1:
Check 18 and 30
18 = 2 X 3 X 3
30 = 2 X 3 X 5
So 18 and 30 are not relatively prime, they have factors in
common
Both numbers can be divided by 2, 3 and 2 X 3 = 6 Of these three
factor numbers the number 6 is the highest Common Factor
(HCF)
Example 3:
Check 140 and 27
140 = 2 X 2 X 5 X 7
27 = 3 X 3 X 3
So 140 and 27 are relatively prime
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Since 6 is not a factor of 411 and 417 , test for 3 or 2
HCF(411,417)= 3
1
...
?

Odd and Even Numbers
Numbers which have 2 as a factor are called Even Numbers,
which do not have 2 as a factor, are called Odd
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An
odd number cannot be divided into two equal part

Multiplication by 5, 50 and 25
Example 1: Find 44 X 5
Multiply by 2 and divide by 2 gives
44 X (5 X 2) / 2 = 44 X10/2
Find 44 X 10 and divide by 2
440/2 = 220
Answer = 220
Example 2: Find 27 X 50
Multiply by 2 and divide by 2 gives
27 X (50 X 2) / 2 = 27 X100/2
Find 27 X 100 and divide by 2
2700/2 = 1350
Answer = 1350
Example
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Table A
mn
11 x 9 = 9 9
12 x 9 = 10 8
13 x 9 = 11 7
--------------18 x 9 = 16 2
19 x 9 = 17 1
20 x 9 = 18 0
Table B
21 x 9 = 18 9
22 x 9 = 19 8
23 x 9 = 20 7

--------------28 x 9 = 25 2
29 x 9 = 26 1
30 x 9 = 27 0
Table C
35 x 9 = 31 5
46 x 9 = 41 4
53 x 9 = 47 7
67 x 9 = 60 3
------------- and so on
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Here LHS of products are uniformly 2 less
than the multiplicands
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Here LHS of products are
uniformly 3 less than the multiplicands
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If
3 is first digit of the multiplicand then LHS of product is 4
less than the multiplicand; if 4 is first digit of the
multiplicand then, LHS of the product is 5 less than the
multiplicand and so on
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