## Types of numbers :

1. Natural numbers (N) = 1, 2, 3, . . . .2. Whole numbers (W) = 0, 1, 2, 3, . . . .

3. Intezers (Z) = −∞ . . . −2, −1, 0, 1, 2, 3, . . .∞

4. Rational numbers (Q) = The numbers of the form

^{p}⁄

_{q }where q ≠ 0. Eg:

^{1}⁄

_{5}, 0.46, 0.333333

5. Irrational numbers ($\mathbb{R} - Q$) = The numbers of the form x

^{1⁄n }≠ Intezer. Also π and e also irrational numbers.

Rational and Irrational numbers together is called Real numbers. It is denoted by $\mathbb{R}$

Other types of numbers:

a. Even numbers : Numbers which are exactly divisible by 2. These numbers are in the format of 2n.

b. Odd numbers: Numbers which gives remainder 1 when divided by 2. These numbers are in the format of 2n ± 1.

c. Prime numbers : The numbers which are divisible by 1 and the number itself are primes. The least prime is 2.

**Note:**There are 15 primes below 50, 25 primes below 100, 168 primes below 1000.

d. Composite numbers : The numbers of which are divisible by more than 2 numbers. First positive composite number is 4.

## Important rules related to Even and Odd numbers:

odd ± odd = even;even ± even = even;

even ± odd = odd

odd × odd = odd;

even × even = even;

even × odd = even.

odd

^{(any number)}= odd

even

^{(any number)}= even

## Fundamental Theorem of Arithmetic:

Any positive integer greater than 1 can be represented as a product of primes only in one way. (Order may be different). Writing a number as a product of primes is called prime factorization format. For example, 100 can be written as 2^{2}× 5

^{2 }in only one way.

##
**Solved Examples**

a. \({\left( {x - z} \right)^2}\) is odd number

b. \(\left( {x - y} \right)z\) is even number

c. \(\left( {z - x} \right){y^2}\) is even number

d. \({\left( {x - y} \right)^2} + z\) is even number

Correct option: d

Solution: Recap the rules related to odd and even numbers.

\(x - z\) = even - odd = odd and odd

^{2}= odd. So option a is correct.

\(\left( {x - y} \right)z\) = (even - even) × odd = even × odd = even. So option b is correct

\(\left( {z - x} \right){y^2}\). We know that y is even. So \({y^2}\) is even. So anything multiplied by even is even. So option c also correct.

\({\left( {x - y} \right)^2} + z\) = even

^{2}+ odd = even + odd = odd. So option d is false.

**Shortcut:**

To solve these type of questions, assume numbers and check options.

Take x = 2, y = 4, z = 3

Clearly, the fourth option = \({\left( {2 - 4} \right)^2} + 3 = 7\) is odd.

2. Let \(x\) be prime and \(y\) be composite. Then

a. \(xy\) is always even

b. \(y - x\) is never even

c. \(\dfrac{{x + y}}{x}\) is never even

d. None of the above

Correct option: d

Explanation: Interesting question. We try to give counter examples to eliminate options one by one.

Take \(x = 3,y = 9\). Then \(xy = 27\) is odd. So option 1 ruled out.

Take \(y = 4,x = 2\). Then \(y - x = 2\) is even. So option 2 ruled out.

Take \(x = 2,y = 6\). Then \(\dfrac{{x + y}}{x} = \dfrac{{2 + 6}}{2} = 4\) is even. So option 3 also ruled out.

So we choose option d.

3. Prime factorization of 12600 is

a. \({2^3} \times {3^2} \times {5^2} \times 7\)

b. \({2^2} \times {3^2} \times {5^2} \times 7\)

c. \({2^2} \times {3^2} \times {5^3} \times 7\)

d. \({2^1} \times {3^1} \times {5^2} \times 7\)

Correct option: a

Explanation:

To find the prime factorization of a number, we continuously divide the given number by prime numbers until the we get prime number. So \(12600 = {2^3} \times {3^2} \times {5^2} \times 7\)

**Important Tip:**The clue in solving this question is, when you observe the number, it is an even number ending with '0'. So it must be divisible by both 2 and 5. So start with 2. After 3 divisions we get 1575 which is clearly divisible by 5. So after 2 divisions we got 63 which is divisible by 7 and 9.

### 4. What least number must be subtracted from 12702 to get number exactly 99 ?

a. 49b. 30

c. 29

d. 31

Correct Option: B

Explanation:

Divide the given number by 99 and find the remainder. If you subtract the remainder from the given number then it is exactly divisible by 99.

99) 12702 (128

__99__

280

__198__

822

__792__

30

Required number is 30.

###

5. The largest number of four digits exactly divisible by 77 is

a. 9768b. 9933

c. 9988

d. 9944

Correct Option: B

Explanation: Find the remainder when 10000 is divided by 77. Then subtract that remainder from 10000.

77) 10000 (12

__77__

230

__154__

760

__693__

67

So 10000 – 67 = 9933 is exactly divisible by 77.

### 6. Sum of the numbers from 1 to 20 is

a. 210b. 110

c. 220

d. 105

Correct Option: A

Explanation: Sum of first n natural numbers = 1 + 2 + 3 + ..... n = $\displaystyle\frac{{n(n + 1)}}{2}$

Substitute n = 20.

So \({S_{20}} = \dfrac{{20 \times 21}}{2} = 210\)

### 7. Sum of even numbers between 15 and 25 is

a. 70b. 80

c. 130

d. 100

Correct Option: D

Explanation: 16 + 18 + .......24.

Taking 2 common we get = 2 ( 8 + 9 + 10 +.....+ 12)

Sum of n natural numbers upto 12

8 + 9 + 10 +.....+ 12 = (1 + 2 + 3 + ......+ 12) - ( 1 + 2 + 3 + ....+ 7)

By applying the formula for the first n natural numbers sum $\displaystyle\frac{{n(n + 1)}}{2}$ we get, $\displaystyle\frac{{12(12 + 1)}}{2} - \displaystyle\frac{{7(7 + 1)}}{2} = 50$

So 16 + 18 + .......24 = 2 × 50 = 100

**Alternative Method:**

16 + 18 + .......24

This is an Arithmetic progression with $a = 16, d = 2, l = 24$

Total terms in the sequence is given by = \(n = \dfrac{{l - a}}{d} + 1\)

So total terms = \(n = \dfrac{{24 - 16}}{2} + 1 = 5\)

Sum of the terms when first term and last term in known = \(\dfrac{n}{2}\left( {a + l} \right)\) = \(\dfrac{5}{2}\left( {16 + 24} \right) = 100\)

### 8. How many numbers between 1000 and 5000 are exactly divisible by 225?

a. 16b. 18

c. 19

d. 12

Correct Option: B

Explanation: First multiple of 225 after 1000 is 1125 (225 × 5 ) and last multiple of 225 before 5000 is 4950 (225 × 22)

Total number are $\displaystyle\frac{{l - a}}{d} + 1 = \displaystyle\frac{{4950 - 1125}}{{225}} + 1 = 18$

### 9. If the first 200 numbers are written down and those divisible by 2 are deleted and again those divisible by 5 are deleted, how many numbers are left out ?

a. 80b. 150

c. 200

d. 160

Correct Option: A

Total numbers divisible by 2 = 100

Total numbers divisible by 5 = 40

But there is double counting. So we have to subtract Total numbers which are divisible by both 2 and 5 i.e, 200 / 10 = 20

So Total numbers which are divisible by either 2 or 5 is 100 + 40 - 20 = 120

Number of numbers which are not divisible by any of those = 200 - 120 = 80

### 10. How many digits are required to write numbers between 1 to 100.

a. 196b. 158

c. 192

d. 200

Correct Option: C

Explanation:

Single digits are from 1 to 9 = 9 digits

Doubt digits are from 10 to 99 = 90 x 2 = 180 digits

100 needs 3 digits. Total 192 digits