Number system is a very important chapter and you will get questions from this area in many competitive exams. We start with classification of numbers.

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

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

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.

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

even ± even = even;

even ± odd = odd

odd × odd = odd;

even × even = even;

even × odd = even.

odd

even

If

\(x.\overline {\underbrace {abc.....}_{n - digits}} \) = \(x.\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\)

Then \(x.\overline {\underbrace {abc.....}_{n - digits}} = x + \dfrac{{abc....}}{{\underbrace {999...}_{n - times}}}\)

\(3.\overline {713} \) =

\(3.\overline {713} \) = \(3 + \dfrac{{713}}{{999}}\) = \(\dfrac{{2997 + 713}}{{999}} = \dfrac{{3710}}{{999}}\)

If

\(x.\underbrace {mnp..}_{k - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\) = \(x.\underbrace {mnp..}_{k - digits}\underbrace {\overline {abc.....} }_{n - digits}\)

In this case, subtract the digits which are not recurring from the whole number and put it in the numerator. In the denominator, put 9's for the number of digits which recur, and put 0's for the number of digits which won't recur.

=\(x.\underbrace {mnp..}_{k - digits}\underbrace {\overline {abc.....} }_{n - digits}\) = \(x + \dfrac{{mnp...abc... - mnp...}}{{\underbrace {999....}_{n - digits}\underbrace {000....}_{k - digits}}}\)

\(12.3\overline {45} \)=

Here only 45 are recurring.

Therefore, \(12.3\overline {45} = 12 + \dfrac{{345 - 3}}{{990}}\) = \(12 + \dfrac{{342}}{{990}}\) = \(12 + \dfrac{{38}}{{110}}\) = \(12 + \dfrac{{19}}{{55}}\) = \(\dfrac{{679}}{{55}}\)

1. Let \(x,y\) are even numbers and \(z\) is an odd number. Then which of the following is false?

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

\(\left( {x - y} \right)z\) = (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

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 \(12600 = {2^3} \times {3^2} \times {5^2} \times 7\)

b. 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

280

822

30

Required number is 30.

a. 9768

b. 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

230

760

67

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

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

b. 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

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 = \(\dfrac{n}{2}\left( {a + l} \right)\) = \(\dfrac{5}{2}\left( {16 + 24} \right) = 100\)

b. 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 $\displaystyle\frac{{l - a}}{d} + 1 = \displaystyle\frac{{4950 - 1125}}{{225}} + 1 = 18$

b. 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

b. 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

## 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, 0.3333335. 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)}= oddeven

^{(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.### Converting recurring decimals into p/q format:

**Model 1:**If

**all the 'n'**digits after the decimal points are recurring, say,\(x.\overline {\underbrace {abc.....}_{n - digits}} \) = \(x.\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\)

Then \(x.\overline {\underbrace {abc.....}_{n - digits}} = x + \dfrac{{abc....}}{{\underbrace {999...}_{n - times}}}\)

**Example:**\(3.\overline {713} \) =

**Solution:**\(3.\overline {713} \) = \(3 + \dfrac{{713}}{{999}}\) = \(\dfrac{{2997 + 713}}{{999}} = \dfrac{{3710}}{{999}}\)

**Model 2:**If

**certain 'n' digits**after the decimal points are recurring, and 'k' digits are not recurring,\(x.\underbrace {mnp..}_{k - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\underbrace {abc.....}_{n - digits}\) = \(x.\underbrace {mnp..}_{k - digits}\underbrace {\overline {abc.....} }_{n - digits}\)

In this case, subtract the digits which are not recurring from the whole number and put it in the numerator. In the denominator, put 9's for the number of digits which recur, and put 0's for the number of digits which won't recur.

=\(x.\underbrace {mnp..}_{k - digits}\underbrace {\overline {abc.....} }_{n - digits}\) = \(x + \dfrac{{mnp...abc... - mnp...}}{{\underbrace {999....}_{n - digits}\underbrace {000....}_{k - digits}}}\)

**Example:**\(12.3\overline {45} \)=

**Solution:**Here only 45 are recurring.

Therefore, \(12.3\overline {45} = 12 + \dfrac{{345 - 3}}{{990}}\) = \(12 + \dfrac{{342}}{{990}}\) = \(12 + \dfrac{{38}}{{110}}\) = \(12 + \dfrac{{19}}{{55}}\) = \(\dfrac{{679}}{{55}}\)

**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