A. Types of Numbers:
1. Natural Numbers: Counting numbers 1, 2, 3, 4, 5,..... are called natural numbers.
2. Whole Numbers: All counting numbers together with zero form the set of whole numbers. Thus,

(i) 0 is the only whole number which is not a natural number.
(ii) Every natural number is a whole number.

3. Integers: All natural numbers, 0 and negatives of counting numbers i.e., {..., - 3 , - 2 , - 1 , 0, 1, 2, 3,.....} together form the set of integers.

(i) Positive Integers: {1, 2, 3, 4, .....} is the set of all positive integers.
(ii) Negative Integers: {- 1, - 2, - 3,.....} is the set of all negative integers.
(iii) Non-Positive and Non-Negative Integers: 0 is neither positive nor negative. So, {0, 1, 2, 3,....} represents the set of non-negative integers, while {0, - 1 , - 2 , - 3 , .....} represents the set of non-positive integers.

4. Even Numbers: A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, etc.
5. Odd Numbers: A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc.
6. Prime Numbers: A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself. Prime numbers up to 100 are: 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.
Prime numbers Greater than 100: Let be a given number greater than 100. To find out whether it is prime or not, we use the following method:
Find a whole number nearly greater than the square root of p. Let k > square root of p. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime. e.g,, We have to find whether 191 is a prime number or not. Now, 14 > square root of 191. Prime numbers less than 14 are 2, 3, 5, 7, 11, 13. 191 is not divisible by any of them. So, 191 is a prime number.
7. Composite Numbers: Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12. Co-primes: Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes.

Example 1: Factor: [latex]4x^{2}[/latex] + [latex]12x[/latex] + [latex]9[/latex]
Solution:

Since [latex]4x^{2}[/latex] = [latex](2x)^{2}[/latex] and 9 = [latex]3^{2}[/latex], we guess [latex](2x + 3)^{2}[/latex]
Test: using the square formula, [latex](2x + 3)^{2}[/latex] = [latex]4x^{2}[/latex] + [latex]12x[/latex] + [latex]9[/latex]
Therefore, [latex](2x + 3)^{2}[/latex] is the answer.

Example 2: Factor: [latex]9x^{2}[/latex] + [latex]24xy[/latex] + [latex]16y^{2}[/latex]
Solution:

Since [latex]9x^{2}[/latex] = [latex](3x)^{2}[/latex] and [latex]16y^{2}[/latex] = [latex](4y)^{2}[/latex], we guess [latex](3x - 4y)^{2}[/latex]
Test: using the square formula, [latex](3x - 4y)^{2}[/latex] = [latex]9x^{2}[/latex] + [latex]24xy[/latex] + [latex]16y^{2}[/latex]
Therefore, [latex](3x - 4y)^{2}[/latex] is the answer.

Example 1: Factor: [latex]x^{2} - 9[/latex]
Solution:

Write [latex]x^{2} - 9[/latex] as [latex]x^{2} - 3^{2}[/latex]
By using the formula, we get [latex](x + 3)(x - 3)[/latex]

Example 2: Factor: [latex]16x^{4} - 81y^{4}[/latex].
Solution:

Write [latex]16x^{4} - 81y^{4}[/latex] as [latex](4x^{2})^{2} - (9y^{2})^{2}[/latex]
By the formula, we get [latex](4x^{2} + 9y^{2})[/latex] [latex](4x^{2} - 9y^{2})[/latex]
Now, [latex](4x^{2} + 9y^{2})[/latex] is a sum of squares (not factorable),
but we can factor [latex](4x^{2} - 9y^{2})[/latex] further as a diffrence of squares again!
Thus, [latex](4x^{2} + 9y^{2})[/latex] [latex](4x^{2} - 9y^{2})[/latex] = [latex](4x^{2} + 9y^{2})[/latex] [latex]((2x)^{2} - (3y)^{2})[/latex]
By the diffrence of squares, we get.
[latex](4x^{2} + 9y^{2})[/latex][latex](2x + 3y)(2x - 3y)[/latex]

Example 1: Factor: [latex]x^{3} + 27[/latex].
Solution:

[latex]x^{3} + 27[/latex] as [latex]x^{3} + 3^{3}[/latex]
By the formula, we get
[latex](x + 3)(x^{2} + 3x + 9)[/latex]

Example 2: Factor: [latex]8x^{3} - 125[/latex].
Solution:

[latex]8x^{3} - 125[/latex] as [latex](2x)^{3} - (5)^{3}[/latex]
By the formula, we get
[latex](2x - 5)(4x^{2} + 10x + 25)[/latex]

• [latex](a+b)^2 = a^2 + b^2 +2ab[/latex]
• [latex](a-b)^2 = a^2 + b^2 -2ab[/latex]
• [latex]a^2 - b^2 = (a+b)(a-b)[/latex]
• [latex](a+b)^2[/latex] - [latex](a-b)^2[/latex] = [latex]4ab[/latex]
• [latex](a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)[/latex]
• [latex](a^3 + b^3) = (a + b)(a^2 - ab + b^2)[/latex]
• [latex](a^3 - b^3) = (a-b)(a^2 + ab + b^2)[/latex]
• [latex](a^3 + b^3 + c^3 - 3abc) = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)[/latex]
• [latex]When \ a + b + c = 0, \ then \ (a^3 + b^3 + c^3) = 3abc[/latex]

1. Find a number such that 20 is subtracted from 9 times the number, the result is 10 more than fourth the number?
Solution:

Let the number to be found be [latex]x[/latex]
When 20 is subtracted from 9 times a number = 9[latex]x[/latex] - 20
The result is 10 more than fourth of the number = 4[latex]x[/latex] + 10
Therefore, 9[latex]x[/latex] - 20 = 4[latex]x[/latex] + 10
⇒9[latex]x[/latex] - 4[latex]x[/latex] = 20 + 10
⇒5[latex]x[/latex] = 30
⇒[latex]x[/latex] = [latex]\frac{30}{5}[/latex]
⇒[latex]x[/latex] = 6
Therefore, the unknown number is 6

2. If the sum of two numbers is 12 and their product is 11. Find the difference between the numbers?
Solution:

Given that sum of two numbers is 12, and product is 11,
Let a + b = 12 and ab = 11
As, [latex](a+b)^2[/latex] - [latex](a-b)^2[/latex] = [latex]4ab[/latex]
By substituting the values,
⇒[latex](12)^2[/latex] - [latex](a-b)^2[/latex] = 4 x 11
⇒144 - [latex](a-b)^2[/latex] = 44
⇒144 - 44 = [latex](a-b)^2[/latex]
⇒[latex](a-b)^2[/latex] = 100
⇒(a - b) = 10
Now consider,
a + b = 12 ----------(i)
a - b = 10 ----------(ii)
By solving equations (i) and (ii)
the two numbers are a = 11 and b = 1
Therefore, difference between the numbers = 11 - 1 = 10

3. If the numbers are added in pairs, the sum equals 12, 17, and 25. Find the numbers?
Solution:

Let the numbers be a, b, and c. Then
Given that
a + b = 12 ---------(i)
b + c = 17 ---------(ii)
c + a = 25 ---------(iii)
By adding equations (i), (ii), and (iii)
a + b + b + c + c + a = 12 + 17 + 25
⇒2(a + b + c) = 54
⇒(a + b + c) = [latex]\frac{54}{2}[/latex]
⇒(a + b + c) = 27 ----------(iv)
Now substitute (i), (ii), and (iii) in (iv)
c = 27 - 12 = 15
a = 27 - 17 = 10
b = 27 - 25 = 2
Therefore, the unknown values are 10, 2 and 15 respectively

4. The average of four consecutive even numbers is 36. Find the smallest and largest number?
Solution:

Let the four consecutive four numbers be
[latex]x[/latex], [latex]x[/latex] + 2, [latex]x[/latex] + 4, [latex]x[/latex] + 6
Given that [latex]\frac{(x + x + 2 + x + 4 + x + 6)}{4}[/latex] = 36
⇒4[latex]x[/latex] + 12 = 36 x 4
⇒4[latex]x[/latex] + 12 = 144
⇒4[latex]x[/latex] = 144 - 12
⇒4[latex]x[/latex] = 132
⇒[latex]x[/latex] = [latex]\frac{132}{4}[/latex]
⇒[latex]x[/latex] = 33
Therefore smallest number is [latex]x[/latex] = 33
Largest number is [latex]x[/latex] + 6 = 33 + 6 = 39

5. The sum of squares of three consecutive odd numbers is 2531. Find the numbers?
Solution:

Let the three consecutive three odd numbers be
[latex]x[/latex], [latex]x[/latex] + 2, [latex]x[/latex] + 4
Given that [latex]x^2 + (x + 2)^2 + (x + 4)^2[/latex] = 2531
[latex]x^2[/latex] + [latex]x^2[/latex] + 4 [latex]x[/latex] + 4 + [latex]x^2[/latex] + 16 + 8[latex]x[/latex] = 2531
⇒3[latex]x^2[/latex] + 20 + 12[latex]x[/latex] = 2531
⇒3[latex]x^2[/latex] + 12[latex]x[/latex] - 2531 + 20 = 0
⇒3[latex]x^2[/latex] + 12[latex]x[/latex] - 2511 = 0
Divide the whole equation by 3
⇒[latex]x^2[/latex] + 4[latex]x[/latex] - 837 = 0
By factorising the equation,
⇒[latex]x^2[/latex] + 31[latex]x[/latex] + 27[latex]x[/latex] - 837 = 0
⇒[latex]x(x + 31)[/latex] - [latex]27(x + 31)[/latex] = 0
⇒[latex](x - 27)[/latex][latex](x + 31)[/latex]
⇒[latex]x[/latex] = 27 or [latex]x[/latex] =-31
Therefore [latex]x[/latex] = 27 (as negative signs cannot be considered, 31 is eliminated)
Hence, the unknown numbers are 27, 29, and 31.