Class 6 – Number-Theory

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Number-Theory
  • Factors and Multiples
  • Divisibility rules
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Number-Theory
  • Square & Square Root
  • HCF
  • LCM
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Introduction: Observe the following pairs of numbers and identify the relation between first number to second number.

i) 12, 36 ii) 15, 3

iii) 12, 18 iv) 16, 5

by observation,

i) 36 can be expressed as product of 12
and 3
⇒ 12 × 3 = 36

∴12 is a factor of 36.

ii) 15 can be expressed as product of 3
and 5

⇒ 15 = 3 × 5

∴ 15 is multiple of 3.

iii) 12 can be expressed as product of 6
and 2, 18 as product of 6, 3

∴ 6 is the common factor of 12 and 18

But there is no direct relation between 12 and 18

iv) 16 cannot be expressed as the product of 5 and any other number.

∴16 and 5 are co-primes (since, 1 is the common factor).

Factor: A number can be expressed as product of two or more numbers, then each of the number in the product is called the factor of the given number.

(or) A number which divides the other number exactly is called the factor of that number (or) A factor of a number is an exact divisor of that number.

Example: 48 = 1 × 48

= 2 × 24

= 3 × 16

= 4 × 12

= 6 × 8

∴ Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

http://images.clipartpanda.com/bloc-clipart-note-md.png Every factor is less than or equal to the given number.

Multiple: If a number is multiplied by 1, 2, 3 …… then the resulting numbers are called multiples of the given number.

Example: Multiples of 4 are 4, 8, 12, 16………

http://images.clipartpanda.com/bloc-clipart-note-md.pngi) Every multiple of a number is greater than or equal to that number.

ii) If a number ‘p’ divides another number ‘q’ exactly, we say that ‘p’ is a factor of ‘q’ and ‘q’ is a multiple of ‘p’.

Example: Show that 2, 3, 5 are factors of 30 also 30 is a multiple of each one of 2, 3, 5.

Solution: 30 =1 × 30

=2 × 15

= 3 ×10

= 5 ×6

∴ Factors of 30 = 1, 2, 3, 5,6, 15, 30

We denote the set of all factors of n by

∴ F(30) = {1, 2, 3, 5, 6, 15,30}

∴ 2, 3, 5 are factors of 30and 30 is the multiple of each one of 2, 3, 5.

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i) 1 is a factor of every number and is the smallest factor.

ii) Every number is a factor of itself and is the largest factor.

iii) Every number is a multiple of itself.

iv) Every number is a multiple of one.

v) The number of multiples of a given number are infinite.

Common factor: When two or more numbers have the same number as factor, it is called a common factor of those numbers.

Example: Find the common factors of 18 and 24.

⇒ Factors of 18 = 1, 2, 3, 6 9, 18

⇒ Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Common factors of 18 and 24 = 1, 2, 3 and 6.

Here, 6 is the highest common factor of 18 and 24.

Even and Odd Numbers:

Even Numbers: All natural numbers exactly divisible by 2(or) all the multiples of 2 are known as even numbers.

Thus, 2, 4, 6, 8, 10, ————— are all even numbers.

Odd Numbers: All natural numbers which are not exactly divisible by 2 (or) which are not the multiples of 2 are called odd numbers.

Thus 1, 3, 5, 7, 9, 11, ——————- are all odd numbers.

http://images.clipartpanda.com/bloc-clipart-note-md.png

i) Two consecutive odd numbers (or) even numbers differ by 2.

ii) Even numbers can be represented as 2n where n is an integer.

iii)Odd numbers can be represented as 2n + 1 (Where n is an integer).

Results to be numbered about even and odd numbers:

Addition and subtraction:

Operation

Result

Example

Even ± Even

Even

4 + 2 = 6, 4 – 2 = 2

Even ± Odd

Odd

6 + 3 = 9, 6 – 3 = 3

Odd ± Even

Odd

7 + 4 = 11, 7 – 4 = 3

Odd ± Odd

Even

9 + 5 = 14, 9 – 5 = 4

Multiplication:

Operation

Result

Example

Even × Even

Even

2 × 4 = 8

Even × odd

Even

4 × 3 = 12

Odd × Even

Even

7 × 6 = 42

Odd × Odd

Odd

5 × 9 = 45

http://images.clipartpanda.com/bloc-clipart-note-md.png Sum of even number of times of odd numbers is even number.

Example: 3 + 5 + 9+ 13 (4 times) = 30 (even)

7 + 15 + 9 + 11 + 3 + 5 (6 times) = 50 (even)

Solved exampleson even and odd numbers:

1. Tell whether 23 + 3 + 5 + 14 + 27 + 18 is even or odd without actual addition.

Solution: 23 + 3 + 5 + 14 + 27 + 18

= O + O + O + E + O + E

= O + O + O + O (4 times) + E + E (2 times)

= E + E

= E.

∴ The final answer is even.

2. Verify(26 × 3) + (14 × 5) + (2 × 8) is even or odd.

Solution: (26 ×3) + (14 × 5) + (2 × 8) = (E × O) + (E × O) + (E × E)

= E + E +E

=E

∴ The final answer is even.

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1) If the product of a certain number of numbers is even, then at least one of them is even

2) If the product of a certain number of numbers is odd, then none of them is even.
3) Sum of first n even natural numbers = n (n +1)

4) Sum of first n odd natural numbers = n2

5) Sum of first n natural numbers =

Prime number: Numbers having exactly two different factors namely, 1 and the number itself are known as prime numbers, and it must be a whole number greater than 1.

Example: The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

http://images.clipartpanda.com/bloc-clipart-note-md.png i) 2 is the only even prime number.

ii)The set of prime numbers are uncountable

iii)2 and 3 are only two primes which are consecutive numbers.

Sieve of Eratosthenes: Eratosthenes was a Greek Mathematician who lived in 200 BC. He developed a simple step by step method for determining which numbers are prime which is called“Eratosthenes Sieve” [A sieve is a strainer that allows small particles to pass through while catching the larger particles].

The following example illustrates how the Sieve of Eratosthenes, can be used to find all the prime numbers that are less than 100.

Step 1. Write the numbers 1 to 100 in ten rows.

Step 2. Cross out 1 because 1 is not a prime.

Step 3. Circle 2 and cross out all multiples of 2. (2,4, 6, 8, 10 …)

Step 4. Circle 3 and cross out all multiples of 3. (3,6, 9, 12, 15 …)

Step 5. Circle 5 and cross out all multiples of 5. (5,10, 15, 20 …)

Step 6. Circle 7 and cross out all multiples of 7. (7,14, 21, 28 …)

Circle all the numbers that are not crossed out and they are the prime numbers less than 100.

∴ Prime numbers less than 100are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71,73, 79, 83, 89 and 97

Composite numbers: Numbers which have more than 2 factors are called composite
numbers.

Example: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 are the first 10 composite numbers.

http://images.clipartpanda.com/bloc-clipart-note-md.pngi) 1is neither prime nor composite.
ii) If x and y are two prime numbers then their product xy will have only 1, x, y and xy as factors.
iii) Any whole number greater than 1 is either prime (or) composite.

Twin primes: If the difference of two prime numbers is ‘2’, then the primes are called ‘Twin primes’.

Example: 3, 5; 5, 7, 11, 13 etc… are twin primes.

http://images.clipartpanda.com/bloc-clipart-note-md.png3, 5,7 is called prime triplet. This is the only triplet in primes with difference2.

Co-Primes: If two (or) more natural numbers which have only ‘1’ as the common factor are called Co-primes.

Example: (2, 3), (3, 4, 5), (4, 9), (8, 13, 15), (16,25) etc.. are co-primes.

http://images.clipartpanda.com/bloc-clipart-note-md.png Co-primes need not be primes themselves.

Example: List all the prime numbers between 1 and 10.

Solution: Prime numbers between 1 and 10 are 2, 3, 5 and 7.

Example: Express (i) 32 as a sum of two odd primes.

(ii) 35 as a sum of three odd primes.

Solution: (i)32 = 3 + 29

(ii) 35 = 5 + 7 + 23.

Example: State whether the numbers 25 and 27 are twin primes or co-primes?

Solution: Factors of 25 = 1, 5, 25

Factors of 27 = 1, 3, 9, 27

∴ 25 and 27 are not prime numbers.

Hence,they are not twin primes.

But both 25 and 27 are having only ‘1’ as the common factor,

∴ 25and 27 are co-primes.

Example: Express 36 as sum of two prime numbers.

Solution: 36 =17+19.

Prime factorization: The process of writing a composite number as the product of prime factors is called prime factorization of the given number.

1) Factor tree method. 2) Short division method.

Factor tree method:

Example: Find the prime factorsof 30.

Solution: First see whether the given number is divisible by least prime number or not.

30 = 2 ´ 15

We have 15 = 3 ´ 5.

So the factors of 30 are 2, 3, 5

For bigger numbers this method takes lot of time. So for bigger number we use

Short division method

Short division method:

Step 1: Divide the given number by the smallest prime factor of the given number.

Step 2: Go on dividing each of the sub sequent quotients by the smallest prime factor, till the last quotient is prime.

Step 3: Express the given number as the product of all these factors.

Example: Express each of following as a product of prime factors:

i) 360 ii)96

Solution:

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You can use a factor tree to list the prime factors. Let us take 102

Use rules of dividing to find any two factors. Continue to find factors until all the factors are prime. The prime factors of 102 are 2, 3 and 17