This lesson contains all the formula regarding to Number system.

2. B.E (final year) Computer Science Engineering, WIT Solapur.
Math is my passion
I like craft, designing and teaching.
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3. NUMBERS
Numbers
Real Numbers Complex (Imaginary)
(a+ib) i=√-1, i²=-1,
Irrational Rational i³= -i , i4= 1
(∏, √2, √3)
Integers Fractions
Whole -ve integers
0 Natural (+ve integers)

4. • Perfect Number:
• Definition : “A perfect number is a number such that the sum of its
factors including 1 and excluding itself is equal to the number ”
Example: 1) Factors of 6 are 1,2,3
1+2+3 = 6
2)Factors of 28 are 1,2,4,7,14
1+2+4+7+14=28
• It is having a standard form 2p−1(2p−1).. Where p is prime number
• Example: perfect numbers are generated by the formula 2p−1(2p−1),
with p a prime number, as follows:
• for p = 2: 21(22−1) = 6
• for p = 3: 22(23−1) = 28
• for p = 5: 24(25−1) = 496
• for p = 7: 26(27−1) = 8128.

5. Classification of Numbers /Integers
• Natural numbers Natural numbers are counting numbers which we use to count . For
example 1,2,3…. The lowest natural number is 1
• Whole numbers When zero is included in the list of natural numbers the numbers
are known as whole numbers .for example , 0, 1, 2….the lowest whole numbers is 0 .
• Integers  Integers are whole numbers ,the negative of whole numbers and zero .for
example ,43434235 ,28,2,0,-28 and -3030 are integers, but numbers like ½ , 2.5 are not
whole numbers .
• Number line  The number line is used to represent the set of real numbers. A
representation of the number line is given below:
• Negative Integers Positive Integers

6. • Real Number:-
• Number which are present on number line are called as Real
Number. Which includes factor, decimal, zeros every possible things.
• Imaginary Number:-
• Number which are not present on number line are called as
imaginary number such number are sq. root of negative.
• e.g. √-9, √-16, √-8
• Solution of these number cannot be see on number line √-1 = I is
imaginary coefficient.
• :. √-9 is 3i, √-16 is 4i

7. • Types of Real Number:-
1) Rational Number :-
These number is present in P/Q form were both P&Q must integer & Q should not be
zero.
E.g. 3/5, -1/2, 4, 10, 3.2
2) Irrational Number:-
• Number which cannot be present in P/Q form are irrational these numbers are non
ending & non repeating.
• e.g. 2.3185….
1.6521…..

8. LCMCONCEPTS
10, 12  60
8, 12  24
14, 10  70
12, 15  60
35 50
7 10 …(Dividing by HCF 5)
70 …(LCM of small number)
70*5=350 …(Re-multiplying by HCF)
Big Number:
84 60
7 5 …(Dividing by HCF 12)
35 …(LCM of small number)
35*12=420…(Re-multiplying by HCF)
Big Number:

9. Square of two digit number
b + 2ab + a
64
a b
b = 16 16 4096
carry 1
2ab = 48 + 48 = 49
carry 4
a = 36 + 36 = 40
2
2
2 2

10. Structure form ( 31 – 80 )
N – 25 d
where,
N is the number
d is the difference between number and 50
For e.g.
1. 54 N = 54 d = 54 – 50 = 4
54 - 25 4 = 16
2. 73 N = 73 d = 73 – 50 = 23
73 – 25 = 48 23 = 529
carry = 5
2
29 16
2
2
53 29

11. STRUCTURE FORM(81 – 100 )
100 – 2d d
where,
N is the number
d is the difference between 100 and number
For e.g.
1. 88 N = 88 d = 100 – 88 = 12
100 – 2(12) = 76 12 = 144
carry 1
2 . 93 N = 93 d = 100 – 93 = 7
100 – 2(7) = 86 7 = 49
2
77 44
86 49
2
2

12. • Method to find out the cubes of two digit numbers :
• The answer has o consist of 4 parts, each of which has to be
calculated separately.
• The first part of the answer will be given by the cube of the ten’s
digit.
• Suppose you have to find the cube of 28.
• The first step is to find the cube of 2 and write it down.
• 2^3 = 8
• The next three parts of the numbers will be derived as follows.
• Derive the values 32, 128 and 512.
• (by creating a G.P. 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 number with its tens digit. In this case the ratio is 8/2 = 4.

13. Now, write the 4 terms in a straight line as below. Also, to the middle two terms
add double the value.
8 32 128 512
+ 64 256
21 9 5 2
(carry over 51)
(8+13) (32+64+43=139) (128+256+51=435)
(carry over 13) (carry over 43)
Hence, 28^3 = 21952

14. 21
8
𝟏
𝟐
4
𝟏
𝟐
2
𝟏
𝟐
1
(1) 8 4
9 2 6 1
34
27 36 48 64
12 72 96
(15) (6)
39 3 0 4

15. For composite number
C= m x n ( Co prime )
( Rule should follow m , n )
Number of divisible /factors
If N is any number which can be factorized like
N= ap *bq *cr * ……
Where a, b, c are prime numbers.
Number of divisors = (p+1)(q+1)(r+1)…….
Q. Find the number of divisors of N =420.
N = 420 = 22 * 31 *71 * 51
So the number of divisors = (2+1)(1+1)(1+1)(1+1) = 24

16. • Number of zeroes in an expression:
• Suppose you have to find the number of zeroes in a product :
• 24*32*17*23*19=(2^3*3^1)*(2^5)*17^1*23*19.
• As you can notice, this product will have no zeroes because it has no 5 in
it.
• However, if you have an expression like : 8*15*23*17*25*22
• The above expression can be the rewritten in the standard form as;
• Zeroes are formed by a combination of 2*5. hence, the number of zeroes
will depends on the number of pairs of 2’s and 5’s that can be formed.
• In above product, there are four twos and two fives. Hence, we shall be
able to form only two pairs of (2*5).
• Hence, there will be 2 zeroes in the product.

17. Process for finding the number of zeroes :
Q. 81!
81/5 = 16
16/5 = 3
Thus, Answer = 16+3 = 19
Q. Find the number of zeroes in 27!
Number of zeroes is 27! = (27/5) + (27/25) where [x] indicates the integer
just lower than the fraction.
Hence, [27/5] = 5 and [27/5^2] = 1+6 = 7 zeroes.
Q. Find the number of zeroes in 137!
[137/5] + [137/5^2] + [137/5^3]
= 27+5+1 = 33 zeroes.

18. • Q. What power of 8 exactly divides 25! ?
• Sol. If were a prime number, the answer should be [25/8] = 3. But since 8
is not prime, use the following process.
• The prime factors of 8 is 2*2*2. For divisibility by 8, we need three twos.
So, every time we can find 3 twos, we add one to the power of 8 that
divides 25! To count how we get 3 twos, we do the following. All even
numbers will give one ‘two’ at least [25/2] = 12.
• Also, all numbers in 25! Divisible by 2^2 will give an additional two
[25/2^2] = 6
• Further, all number in 25! Divisible by 2^3 will give a third two. Hence
[25!/2^3] = 3
• And all numbers of twos in 25! Divisible by 2^4 will give a fourth two.
Hence [25!/2^4] = 1
• Hence, total number of twos in 25! Is 22. For a number to be divided by 8,
we need three twos. Hence, since 25! has 22 twos, it will be divided by 8
seven times.

19. PRIME
NUMBER
Prime No. < 100 :
2, 3, 5, 7, 11, 13,
17, 19, 23, 29, 31,
37,41, 43, 47, 53,
59, 61, 67, 71, 73,
79, 83, 89, 97
To check 359 is a
Prime no. or not
Sol: Approx. sq.
root = 19
Prime no. < 19 : 2,
3, 5, 7, 11, 13, 17
There are 15
prime no. from 1
to 50
and 25 prime no.
from 1 to 100
There are 168
prime no.
from 1 to 1000

20. • Divisibility Rules 
• For 2 : If unit digit of any number is 0,2,4,6 or 8, then that number will be divisible by 2
• For 3 : If sum total of all the digits of any numbers is the visible by 3 then the number will be
divisible by 3, (e g ; 123 ,456 etc )
• For 4 : If the last two digits of a numbers are divisible by 4 , then that number will be divisible
by 4 (e g ; 3796 ,248 etc )
• For 5 : If the last digit of the number is 5 or 0 , then that number will be divisible by 5
• For 6 : Rule should follow 2 , 3
• For 7 : ( R -2L) should be multiple of 7
• R  Remaining number
• L  Last digit
• e g :- 441 , here R = 44 & L = 1 hence ,
• R- 2L= 44 – 2(1)
• =42

21. • For 8  If the last 3 digits of the number are divisible by 8, then the number itself will
be divisible by 8 .
• E g ;128, 34568 etc
• For 9  If the sum of the digit of the numbers is divisible by 9 .
• E g; 129835782
• For 11  The difference between the sum of the digit at the even places and the sum
of the digits at the odd places is divisible by 11/0
• e g ; 6595149 is divisible by 11
• 6+9+1+9 (odd) = 25 and
• 5+5+4 (even) = 14
• For 12  divisible by 3 & 4
• For 13  (R + 4L)
• e g ; 1404
• R = 140 L= 4
• 140 +16 = 156 (13n)
• For 14  both by 2 & 7
• For 15  by 5 & 3