Introduction to Number System

Introduction to Number System

Number System

 

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.

 

Types of numbers :

1. Natural numbers (N) = 1, 2, 3, . . . .

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

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

4. Rational numbers (Q) = The numbers of the form \(\frac{p}{q}\) where q ≠ 0.  Eg:\(\frac{1}{5}\) , 0.46, 0.333333

5. Irrational numbers (R−Q) = The numbers of the form \(x^{\frac{1}{n}}\)  ≠ Integer.  Also π and e also irrational numbers.

Rational and Irrational numbers together is called Real numbers. It is denoted by 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.

Prime Numbers from 1 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.

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

e. Positive, Negative and Zero: Any number more than zero is positive, while any number less than zero is negative. Zero is neither positive nor negative.

 

Note:

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

1 is neither prime nor composite.

The smallest prime number is 2.

2 is the only even prime number; remaining all prime numbers are odd.

 

Procedure to find out the prime number:

Suppose A is given number.

Step 1: Find a whole number lets say “K’ nearly greater than the square root of A. K > \(\sqrt{A}\)

Step 2: Test whether A is divisible by any prime number less than K. If yes A is not a prime number. If not, A is prime number.

 

Example 1:

Find out whether 337 is a prime number or not?

Step 1: 19 > \(\sqrt{337}\). Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17

Step 2: 337 is not divisible by any of them. Therefore 337 is a prime number.

 

Example 2:

Find out whether 221 is a prime number or not?

Step 1: 15 > \(\sqrt{221}\). Prime numbers less than 15 are 2, 3, 5, 7, 11, 13

Step 2: 221 is divisible by 13. Therefore 221 is not a prime number.

Important rules related to Even and Odd numbers:

Addition and Subtraction: odd ± odd = even; even ± even = even; even ± odd = odd;

 

Multiplication: odd × odd = odd; even × even = even; even × odd = even;

 

Powers: \(odd^{{}\,\,{any \,\, number} \,}\)= odd ; \(even^{{}\,\,{any \,\, number} \,}\)= even;

 

Note:

Adding/subtracting two similar terms (either both are even or odd) will always give output as even result.

Even always multiplied by any integer gives output as even result.

For product of integers to be odd, all multiplying terms have to be odd.

The result of exponential expression is same as the base.

We can’t generalise the rules for division.

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 \(2^{2}\times 5^{2}\)  in only one way.

 

Converting recurring decimals into \(\frac{p}{q}\) format:

Model 1:

If all the 'n' digits after the decimal points are recurring, say,

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

 

Then \(x.\underbrace{\overline{abc...............}} = x + \underset{n-times}{\underbrace{\frac{abc..........}{999.........}}}\)

 

Example:

\(5.\overline{374} = 5 + \frac{374}{999} = \frac{4995+374}{999} = \frac{5369}{999}\)

 

Model 2:

If certain 'n' digits after the decimal points are recurring, and 'k' digits are not recurring,

 

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

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

 

Example:

\(15.5\overline{96} = 15 + \frac{596-5}{990} = 15 + \frac{591}{990} = 15 + \frac{197}{330} = \frac{5147}{330}\)

 

Important points to remember:

Properties of zero: It is a real number, but neither positive nor negative.

It is a rational number.

It is an even number.

It is a whole number, but not a natural number.

1 is neither prime nor composite.

2 is an even prime number, which is a smallest prime number.

\(\Pi\) is a irrational number. \(\frac{22}{7}\) is rational number.


 

Solved Examples

1.  Let a,b are even numbers and c is an odd number.  Then which of the following is false?

a. \((a-c)^{2}\) is odd number

b. (a−b)c is even number

c. (c−a)\(b^{2}\) is even number

d. \((a-b)^{2}\) + c is even number

 

Correct option: d

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

a-c = even - odd = odd and \(odd^{2}\) = odd. So option a is correct.

(a−b)c = (even - even) × odd = even × odd = even.  So option b is correct

(c−a)\(b^{2}\).  We know that b is even. So \(b^{2}\) is an even. So anything multiplied  by even is even. So option c also correct.

\((a-b)^{2}\) + c =\(even^{2}\) + odd = even + odd = odd. So option d is false.

 

Shortcut:

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

Take a = 2, b = 4, c = 3

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

 

2. Given a, b, and c are odd integers. If s=a(a+b+c)+b(a+b+c)+c(a+b+c). Which of the following could be true?

I. \(s^{2}\) -2s+1 is odd.  II. \(s^{2}\) +2s+1 is even.  III. s=0.    IV. s<0.    V. (s+1)(s-1) is even.

 

a. I, II, and IV

b. I, II, and V

c. II, IV and V

d. II, V

 

Correct option: d

 

Solution: s=a(a+b+c)+b(a+b+c)+c(a+b+c)=(a+b+c)(a+b+c)=\((a+b+c)^{2}\)

Recap the rules related to odd and even numbers.

s=\((odd+odd+odd)^{2}\) = \((odd)^{2}\) = odd

I. \(s^{2}\) - 2s+1 = \((s-1)^{2}\) =\((odd-1)^{2}\) = \((even)^{2}\) = even. (false)

II. \(s^{2}\) +2s+1 = \((s+1)^{2}\) = \((odd+1)^{2}\) = \((even)^{2}\) =even (true)

III. s=0 (s is odd number. So, it can’t be even) (false)

IV. s<0 (s is square of number. So, it can’t be negative) (false)

V. s+1 = odd+1=even, s-1=odd-1=even. (s+1)(s-1) = (even)(even)=even. (true)

 

3. Let a be prime and b be composite. Then which of the following options is correct?

 

a. ab is always even

b. b−a is never even

c. \(\frac{{a + b}}{a}\) is never even

d. None of the above

 

Correct option: d

Solution: Interesting question.  We need to give counter example to eliminate options one by one.

Take a=3,b=9.  Then ab=27 is odd. So option 1 ruled out.

Take b=4,a=2.  Then b−a=2 is even. So option 2 ruled out.

Take a=2,b=6.  Then \(\frac{{a + b}}{a}\)=\(\frac{{2 + 6}}{2}\)=4 is even. So option 3 also ruled out.

So we choose option d.

 

4.  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

 

Solution:

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

 

prime factorization

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.

 

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

a. 49

b. 30

c. 29

d. 31

 

Correct Option: B

Solution:

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.

 

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

a. 9768

b. 9933

c. 9988

d. 9944

 

Correct Option: B

Solution:  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.

 

7. Sum of the numbers from 1 to 40 is

a. 820

b. 910

c. 1220

d. 1050

Correct Option: A

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

Substitute n = 40.

So \({S_{40}}\) = \(\frac{{40 \times 41}}{2}\) = 820

 

8. Find the number of even numbers from 1 to 100 is

a. 50

b. 100

c. 25

d. 51

Correct Option: A

Solution: Let us count number of even numbers from 1 to 10 i.e 2, 4, 6, 8, 10.

So, Half of the numbers are even numbers.

From 1 to 100 the number of even numbers are 50.

Alternative Method:

From  1 to 100, first number even is 2, and Last number even is 100.

2, 4, 6, 8, 10, 12…………….98, 100

This is an Arithmetic progression with a=2, d=2 , l=100

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

So total terms = n=\(\frac{{100 - 2}}{2} + 1\) = \(\frac{{98}}{2} + 1\) = 49+1=50

 

9. Find the sum of even numbers from 100 to 200 is

a. 7650

b. 5670

c. 2550

d. 5140

Correct Option: A

Solution: From  100 to 200, first number even is 100, and Last number even is 200.

100, 102, 104, 106, 108, 110, 112…………….198, 200

This is an Arithmetic progression with a=100 , d=2 , l=200

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

So total terms = n=\(\frac{{200 - 100}}{2} + 1\) = \(\frac{{100}}{2} + 1\) = 50+1=51

Sum of the terms when first term and last term in known =\(\frac{n\times(a + l)} {2}\)\(\frac{51\times( 100+ 200)} {2}\) = 7650

10. Find the number of multiples of 3 from 1 to 100 is

a. 50

b. 33

c. 25

d. 51

Correct Option: B

Solution: The number of multiples of 3 from 1 to 100 is \(\frac{100}{3}\) = 33.33.

We need to consider an integer that is less than or equal to that which is a whole number.

So, the number of multiples of 3 from 1 to 100 is 33.

Alternative Method:

From  1 to 100, first number multiple of 3 is 3, and Last number multiple of 3 is 99.

3, 6, 9, 12…………….96, 99

This is an Arithmetic progression with a=3, d=3, l=99.

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

So total terms = n=\(\frac{{99 - 3}}{3} + 1\) = \(\frac{{96}}{3} + 1\) = 32+1=33

 

11. Find the number of multiples of 5 from 1 to 200 is

a. 50

b. 40

c. 41

d. 51

Correct Option: B

Solution: The number of multiples of 5 from 1 to 200 is \(\frac{200}{5}\) = 40.

We need to consider an integer that is less than or equal to that which is a whole number.

So, the number of multiples of 5 from 1 to 200 is 40.

Alternative Method:

From  1 to 200, first number multiple of 5 is 5, and Last number multiple of 5 is 200.

5, 10, 15, 20…………….195, 200

This is an Arithmetic progression with a=5, d=5, l=200.

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

So total terms = n=\(\frac{{200 - 5}}{5} + 1\) = \(\frac{{195}}{5} + 1\) = 39+1=40

 

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

a. 16

b. 18

c. 19

d. 12

Correct Option: B

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

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

 

13. How many numbers from 1 to 300 which are exactly divisible by both 5 & 6?

a. 70

b. 60

c. 10

d. 50

Correct Option: C

Solution:

Total numbers which are divisible by both 5 and 6 are divisible by 30 i.e, \(\frac{300}{30}\) = 10

Note: Divisible by both means, we need to find LCM of (5, 6) i.e 30.

 

14. How many numbers from 1 to 50 which are exactly divisible by 4 but not 5?

a. 10

b. 12

c. 14

d. 24

Correct Option: A

Solution:

The number of multiples of 4 from 1 to 50 are: {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}

The numbers from multiples of 4 which are also multiples of 5 are {20, 40}

The numbers of multiples of 4 but not 5 are : {4, 8, 12, 16, 24, 28, 32, 36, 44, 48}

Total numbers divisible by 4 = 12

Total numbers divisible by 4 and 5 = 2

Total numbers divisible by 4 but not 5 = 12-2 = 10

Alternative Method:

Total numbers divisible by 4 = \(\frac{50}{4}\)=12

Total numbers divisible by 4 and 5 = \(\frac{50}{20}\)=2

Total numbers divisible by 4 but not 5 = 12-2 = 10

Shortcut:

15. How many numbers from 1 to 100 which are exactly divisible either 2 or 3?

a. 67

b. 83

c. 66

d. 50

Correct Option: A

Solution:

Total numbers divisible by 2 = 50

Total numbers divisible by 3 = 33

But there is double counting. So we have to subtract Total numbers which are divisible by both 2 and 3 i.e, \(\frac{100}{6}\) = 16

So Total numbers which are divisible by either 2 or 3 is 50 + 33 - 16 = 67

 

Shortcut:

34+16+17 = 67

 

16. 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. 80

b. 150

c. 200

d. 160

Correct Option: A

Solution:

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, \(\frac{200}{10}\) = 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

 

Shortcut:

200-120=80

 

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

a. 196

b. 158

c. 192

d. 200

Correct Option: C

Solution:

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

 

18. How many digits are required to write numbers between 1 to 5000.

a. 19145

b. 22786

c. 25169

d. 18893

Correct Option: D

Solution:

Single digits are from 1 to 9 = 9 digits

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

Triple digits are from 100 to 999 = 900 x 3 = 2700 digits

Four digits are from 1000 to 5000 = 4001 x 4 = 16004

Total digits used from 1 to 5000 are 9 + 180 + 2700 + 16004 = 18893

19. In writing all of the integers from 1 to 300, how many times is the digit 1 used?

a. 300

b. 60

c. 100

d. 160

Correct Option: D

Solution:

Let's consider the values from 000 to 299

NOTE: Yes, I have started at 000 and ended at 299, even though though the question asks us to look at the numbers from 1 to 300.

HOWEVER, notice that 000 and 299 do not have any 1's so the outcome will be the same.

 

Okay, there are 300 integers from 000 to 299

 

UNITS position

Each of the 10 digits (0, 1, 2, 3..., 9) appears the same number of times in the UNITS position.

So, the digit 1 must occur \(\frac{1}{10}\) of the time in this position.

So, the digit 1 must appear \(\frac{1}{10}\)\(\times\)(300), which equals 30 times.

 

TENS position

Each of the 10 digits (0, 1, 2, 3..., 9) appears the same number of times in the TENS position.

So, the digit 1 must occur \(\frac{1}{10}\) of the time in this position.

So, the digit 1 must appear \(\frac{1}{10}\)\(\times\)(300), which equals 30 times.

 

HUNDREDS position

The 1 appears in the HUNDREDS position in every integer from 100 to 199

So, the digit 1 appears 100 times.

 

TOTAL = 30 + 30 + 100

= 160

 

20. How many integer solutions for both x and y are there for the equation 2x+4y=17?

a. No Solution

b. 2

c. 1

d. 17

Correct Option: A

Solution:

If x and y are integers then 2\(\times\)(integer) +4\(\times\)(integer) = even + even = even.

So, 2x+4y must be an even number.

17 is an odd number.

No solution.