The number system is the basis of entire basic numeracy. Most of its content is already been studied at school level. Almost onefifth of every competitive examination questions are from this chapter, so it is expected from the students to prepare this chapter well.
Classification of Numbers
There are infinite natural numbers and 1 is the smallest natural number. On the basis of divisibility, there could be two types of natural numbers: Prime and Composite.
Properties of Natural Numbers
Even Numbers: All natural numbers divisible by 2 are even numbers, e.g., 2, 6, 10, 100, 1206, …
Represented as ‘2n’, where n is natural number.
Odd Number: All natural numbers which are not divisible by 2 are odd numbers. E.g., 1, 3, 5, 7, ….
Represented as ‘2n + 1’, where n is natural number.
Prime Numbers: a number that is divisible only by itself and unity. E.g. 2, 3, 5, 7, 11, 13, 17, 19, ….
Some Properties of Prime Numbers:
 The smallest prime number is 2 and is the only even prime number.
 The lowest odd prime number is 3
 There are total 25 prime numbers upto 100
 There are total 46 prime number upto 200
 1 is neither prime nor composite number.
 There are infinite prime numbers.
 List of Prime numbers from 1 to 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.
How to check for a Prime Number?
To check whether a number ‘N’ is prime or not, adopt the following method.
a) Take the square root of the number.
b) Round of the square root to the immediately lower integer. Call this number ‘z’. E.g., If you want to check for 157, its square root will be 12.
c) Check the divisibility of the number N by all the prime numbers below the ‘z’. If there is no prime number below the value of ‘z’ which divides N then the number is prime.
E.g., Check whether 157 is prime or not.
√157 = 12 (Lower Integer)
Prime numbers below 12 are: 2,3,5,7,11.
Since 157 is not divisible by any of them it is a prime number.
For higher numbers application of rules of divisibility and properties of squares of even and odd numbers are used to save time while predicting a number as prime or not.
Some rules for Coprimes:
For two numbers to be prime:
 Two consecutive natural numbers are always coprime (E.g. 5, 6; 82, 83 and so on)
 Two consecutive odd numbers are always prime (E.g. 7, 9; 51, 53; and so on)
 Two prime numbers are always coprime (E.g. 13, 17; 53, 71 and so on)
 One prime number and another composite number (such that the composite number is not a multiple of that prime number) are always co prime (E.g. 17, 38; 23, 49 and so on)
Rules for spotting 3 coprime numbers:
 Three consecutive odd numbers are always coprime (E.g. 15, 17, 19)
 Three consecutive natural numbers with the first one being odd (E.g. 15, 16, 17; 41, 42, 43 and so on) are coprime.
 Two consecutive natural numbers along with the next odd number (E.g. 22, 23, 25; 68, 69, 71 and so on) are coprime.
 Three prime numbers (E.g. 17, 23, 29)
Factors and Multiples
Factors: A number is said to be the factor of another number if divides the other exactly. Thus, 8 and 5 are factors of 40.
Common Factors: A common factor of two or more numbers is a number that divides each of them exactly. Thus, 3 is a common factor of 9, 12, 27.
Multiples: Multiple of any number ‘k’ is represented by kn, where n is natural number. E.g., multiples of 2 are 2,4,6,8,10, …
Common Multiple: A common multiple of two or more numbers is a number which is exactly divisible by each of them. Thus, 30 is a common multiple of 2,3,5,6,10 and 15.
Thus a number which divides a given number exactly is called factor ( or divisor) of that given number and the given number is called multiple of that number.
e.g., 30 is exactly divisible by 1,2,3,5,6,10,15 and 30. Therefore these are called its factors of 30 while 30 is called as the multiple of these factors. 1 and 30 are called improper factors and 2,3 and 5 are called proper factors.
Points to note:
 1 is factor of every number.
 Every number is a factor of itself.
 Every number, except 1 has at least two factors viz., 1 and itself.
 Every factor of a number is less than or equal to that number.
 Every multiple of a number is greater than or equal to itself.
 Every number has infinite number of its multiples.
 Every number is a multiple of itself.
Prime Factorisation: It is the method of expressing a number into its prime factors.
e.g., 30 = 2 x 3 x 5
210 = 2 x 3 x 5 x 7
Perfect number: is a number of which sum of its factors (excluding itself) is equal to the number itself. E.g., 6 (sum of its factors 1,2 and 3 is 6).
Number of Factors
How to find the Number of Factors?
Number of Factors: Any given number ‘N’ can be represented as
N = a^{p }.b^{q} .c^{r}….
Where a, b and c are its prime factors and p, q and r are the power of its prime factors.
Therefore Number of factors of N is given by (p + 1)(q + 1)(r + 1)…..
Q) Find the number of factors of 40.
Sol) Represent 40 in its prime factors
40 = 2 x 2 x 2 x 5
= 2^{3} x 5^{1}
Number of factors of 40 = (3 + 1)(1 + 1) = 4 x 2 = 8; here p is 3 and q is 1.
To check: number of factors are: 1, 2, 4, 5, 8, 10, 20 and 40 (Total 8)
Explanation (why 1 is added in number of factors?) 
The proper expansion of the expression ‘N = a^{p }.b^{q} .c^{r}….’ is (a^{0 }+ a^{1} + a^{2} + a^{3} + …. + a^{p})(b^{0} + b^{1} + b^{2} + … + b^{q})(c^{0} + c^{1} + c^{2} + …. c^{r});
Hence,
the term with power 0 which is nothing but unity leads to add 1 while counting the number of factors. 
Number of Odd Factors
The expression N = a^{p }.b^{q} .c^{r}…. contains only one even prime number (i.e., 2) in term a^{p }, so while calculating the number of odd factors we only take the term a^{0 }from even factors. Thus, the formula to calculate the Number of Odd Factors is:
Number of Odd Factors =
(1).(q + 1).(r + 1).(s + 1)…..
Or
(q + 1).(r + 1).(s + 1)….
where p, q, r, …. are odd factors
Q) Find the number of odd factors of 24.
Sol) Prime factorization of 24 is
24 = 2^{3} x 3^{1 }
Number of odd factors are (q + 1) = (1 + 1) = 2
Q) Find the number of odd factors of 36.
Sol) 36 = 2^{2 }x 3^{2 }
Number of odd factors are (2 + 1) = 3 viz., 1, 3, 9
Q) Find total number of factors and the number of odd factors of 180.
Sol) 180 = 2^{2} x 3^{2} x 5^{1 }
Total number of factors = (2 + 1)(2 + 1)(1 + 1) = 18
Number of odd factors are (2 + 1)(1 + 1) = 3 x 2 = 6
Number of Even Factors
From the expression N = a^{p }.b^{q} .c^{r}…. of prime factorization of any number.
The number of even factors = (p)(q + 1)(r + 1)….
Q) Find the number of even factors of 100.
Sol) 100 = 2^{2} x 5^{2 }
Number of even factors = (2)(2 + 1) = 2 x 3 = 6
Q) Find the total, odd and even number of factors of 360.
Sol) 360 = 2^{3} x 3^{2} x 5^{1}
Total number of factors = (3 + 1)(2 + 1)(1 + 1) = 4 x 3 x 2 = 24
Total number of odd factors = (2 + 1)(1 + 1) = 3 x 2 = 6
Total number of even factors = (3)(2 + 1)(1 + 1) = 3 x 3 x 2 = 18
Note: To save time find total number of factors and odd, subtract odd from total factors to get the number of even factors. As in above case total number of factors is 24 and odd number of factors are 6. Thus 24 – 6 = 18 which is the number of even factors of 360.
Sum of Factors (Odd and Even)
From the expression of prime factorisation of any number N,
N = a^{p }.b^{q} .c^{r}….
Sum of factors = (a^{0} + a^{1} + a^{2} + … + a^{p})(b^{0} + b^{1} + b^{2} + … + b^{q})(c^{0} + c^{1} + c^{2} +… +c^{r})…
Sum of odd factors = (b^{0} + b^{1} + b^{2} + … + b^{q})(c^{0} + c^{1} + c^{2} +… +c^{r})…
Sum of even factors = (a^{1} + a^{2} + … + a^{p})(b^{0} + b^{1} + b^{2} + … + b^{q})(c^{0} + c^{1} + c^{2} +… +c^{r})…
Note: In case of sum of even factors we don’t count the a^{0 }term and start with a^{1} term.
Q) Find the sum of factors, sum of odd and even number of factors of 240.
Sol) 240 = 2^{4} x 3^{1} x 5^{1 }
The sum of factors will be given by:
(2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4})(3^{0} + 3^{1})(5^{0} + 5^{1})
= (1 + 2 + 4 + 8 + 16)(1 + 3)(1 + 5) = 31 x 4 x 6 = 744
Sum of odd number of factors will be given by:
(3^{0} + 3^{1})(5^{0} + 5^{1})
= 4 x 6 = 24
Sum of even number of factors will be given by:
(2^{1} + 2^{2} + 2^{3} + 2^{4})(3^{0} + 3^{1})(5^{0} + 5^{1})
= (2 + 4 + 8 + 16)(1 + 3)(1 + 5) = 30 x 4 x 6 = 720
Note: To save time, find the sum of total number of factors and then the sum of odd factors. From the sum of total factors subtract the sum of odd to get the sum of even numbers of factors.
As in the above case total sum is 744 and the sum of odd factors is 24. Thus 744 – 24 = 720 which is the sum of even numbers of factors.
Q) Find the sum and the number of factors of 1200 such that the factors are divisible by 15.
Sol) 1200 = 2^{4} x 5^{2} x 3^{1 }
For
a factor to be divisible by 15 it should be compulsorily have 3^{1} and
5^{1} in it.
Thus, sum of factors divisible by 15 = (2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4})(5^{1} + 5^{2})(3^{1}) and consequently the number of factors will be given by = 5 x 2 x 1 = 10.
Number of ways of expressing a composite number as a product of two factors = ½ the no. of total factors
E.g., 24 = 1 x 24
2 x 12
3 x 8
4 x 6
And total number of factors of 24 = 2^{3 }x 3^{1 }= (3 + 1) (1 + 1) = 4 x 2 = 8
And ½ of 8 = 4. Therefore, 4 is the number of ways 24 is expressed as a product of two numbers.
Average of Factors = Sum of Factors/number of factors
Total number of Prime Factors:
From expression N = a^{p }.b^{q} .c^{r}….
Number of Prime Factors are: p + q + r
Successive Division
If the quotient in a division is further used as a dividend for the next divisor and again the latest obtained divisor is used as a dividend for another divisor and so on, then it is called the Successive Division.
Q) Find the least possible number which when divided by 2, 5, 4 and 3 leaves the remainder 1, 1, 3 and 1 respectively.
Sol) Arrange the divisors and remainders, as remainders just below the divisors
Divisors: 2 5 4 3
Remainders: 1 1 3 1
Now start from the right side (i.e. 3 in above case) and add while going down and multiply while going above left.
Step 1: (3 + 1) x 4 = 16
Step 2: (16 + 3) x 5 = 95
Step 3: (95 + 1) x 2 = 192
Step 4: 192 + 1 = 193
So, the least such possible number is 193.
And all higher such numbers are: (2 x 5 x 4 x 3) + 193
120m + 193 ; where m = 1, 2, 3……
Fractions and Decimal Fractions
If any unit be divided into any number of equal parts, one or more of these parts is called a fraction of the unit. The fraction onefifths, twothirds, threefourths are written as: 1/5, 2/3, 3/4 respectively.
The lower number, which indicates the number of equal parts into which the unit is divided, is called denominator.
The upper number, which indicates the number of parts taken to form the fraction, is called the numerator.
The numerator and the denominator of a fraction are called its terms.
Testing
When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are
\[x = {b \pm \sqrt{b^24ac} \over 2a}.\]
\[\displaystyle
\text { Simplify: } 1+\frac{1}{1+\frac{2}{2+\frac{3}{1+\frac{4}{5}}}}
\]
\(
\text { Simplify: } \displaystyle{1+\frac{1}{1+\displaystyle{\frac{2}{2+\displaystyle{\frac{3}{1+\displaystyle{\frac{4}{5}}}}}}}}
\)
Simplify: \(1+\dfrac{1}{1+\dfrac{2}{2+\dfrac{3}{1+\dfrac{4}{5}}}}\)
\[
\text { Find the value of } \frac{2}{1+\frac{1}{1\frac{1}{2}}} \times \frac{3}{\frac{5}{6} \text { of } \frac{3}{2} \div 1 \frac{1}{4}}
\]
Find the value of \(\frac{2}{1+\frac{1}{1\frac{1}{2}}} \times \frac{3}{\frac{5}{6} \text { of } \frac{3}{2} \div 1 \frac{1}{4}}\)
If \(x^{2}7 x+1=0\), then find the value of :
\[\frac{20 x}{5 x^{2}15 x+5}\]
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
\[x^{2}7 x+1=0\]
\[\Rightarrow x^{2}+1=7 x\]
\[\Rightarrow \frac{20 x}{5 x^{2}15 x+5}=\frac{20 x}{5\left(x^{2}+1\right)15 x}\]
\[\Rightarrow \frac{20 x}{5 \times 7 x15 x}=\frac{20 x}{35 x15 x}=\frac{20 x}{20 x}=1\]
\begin{aligned}
\frac{17}{7+\frac{3}{4\frac{11}{4}}} & \times \frac{2021}{2193} \div\left(\frac{85}{48}\frac{15}{16}\right)+\frac{3}{4} \text { of } \frac{\frac{3}{2} \frac{2}{\frac{5}{2}}}{\frac{17}{2}} \\
&=\frac{17 \frac{3}{1611}} \times \frac{47}{51} \div\left(\frac{85}{48}\frac{15}{16}\right)+\frac{3}{4} \text { of } \frac{3 \frac{3}{4}}{\frac{5}{2}} \\
&=\frac{17}{7+\frac{12}{5}} \times \frac{47}{51} \div\left(\frac{8545}{48}\right)+\frac{3}{4} \times \frac{15}{4} \times \frac{2}{5} \\
&=\frac{17}{\frac{35+12}{5}} \times \frac{47}{51} \div \frac{40}{48}+\frac{9}{8} \\
&=\frac{17 \times 5}{47} \times \frac{47}{51} \times \frac{48}{40}+\frac{9}{8} \\
& 2+1 \frac{1}{8}=3 \frac{1}{8}
\end{aligned}
\[
R=\rho \frac{l}{A}=\frac{l}{K A}
\]
\(1 / \mathrm{A}=\) cell constant or \(\mathrm{G}^{*}\)
\[
\begin{array}{c}
\mathbf{G}^{*}=\frac{l}{A}=R K \\
K=\frac{G^{*}}{R}
\end{array}
\]
Molar Conductivity \(\left[\Lambda_{m}\right]\)
\[
\wedge_{m}=\frac{K}{C}
\]
When the concentration of a electrolyte reaches zero, the molar conductivity is called limiting molar conductivity denoted by \(\wedge_{m}^{\circ}\)
For strong electrolyte
\[
\wedge_{m}=\wedge_{m}^{\circ}=A c^{\frac{1}{2}}
\]
According to Kohlraush, difference in \(\wedge_{m}^{\circ}\) of the electrolyte \(\mathrm{NaX}\) and \(\mathrm{KX}\) for any \(\mathrm{X}\) is nearly constant.
Also,
\[
\begin{array}{c}
\Lambda_{m(N a C l)}^{\circ}=\lambda_{N a^{+}}^{0}=\lambda_{C l^{}}^{0} \\
\wedge_{m}^{\circ}=v_{+} \lambda_{+}^{\circ}=v_{} \lambda_{}^{0}
\end{array}
\]

1) How to check for a Prime Number?
Ans) To check whether a number ‘N’ is prime or not, adopt the following method. a) Take the square root of the number. b) Round of the square root to the immediately lower integer. Call this number ‘z’. E.g., If you want to check for 157, its square root will be 12. c) Check the divisibility of the number N by all the prime numbers below the ‘z’. If there is no prime number below the value of ‘z’ which divides N then the number is prime. E.g., Check whether 157 is prime or not. √157 = 12 (Lower Integer) Prime numbers below 12 are: 2,3,5,7,11. Since 157 is not divisible by any of them it is a prime number. Note: For higher numbers, application of  rules of divisibility and properties of squares of even and odd numbers are used to save time while predicting a number as prime or not.
Did You Know
Perfect number: is a number in which sum of its factors (excluding itself) is equal to the number itself. E.g., 6 (sum of its factors 1, 2 and 3 is 6)