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Greatest Common Divisor - Mathematics Form 1

GREATEST COMMON DIVISOR (G.C.D)

BACKGROUND KNOWLEDGE

Find out common factors (Common divisors)

Factor - If a number can be divided by another number

without a remainder, the second number is a factor of the first.

Worked out examples

Find the divisors/ factors of the following set of numbers;


1. 20, 32


2. 36, 48, 60



3. Divisibility test for prime numbers, that is 2,3,5,11

A number is divisible by:

2 - If the last digit is zero or even e.g 30,102, 6, 76

3 - If the sum of the digits are divisible by 3 ,

for example

1 2 3 .........{1+ 2 + 3 = 6}

3 4 5 ........ { 3 + 4 + 5 = 12}

2 0 4 .........{ 2 + 0 + 4 = 6}

5- If the last digit is 0 or 5 e.g. 50, 65, 70

11 - If the last sum of the alternate digit are equal

or their difference is a multiple of 11,

e.g 1 8 7 ...... ( 1 +7) - 8= 0

8 1 9 3 9....{8 + 9 + 9} -{1 +3} = 22

Lesson Objective(s)

By the end of this lesson you should be able to;

apply G.C.D to real life situations appropriately.

GREATEST COMMON DIVISOR (G.C.D)

Introduction


G.C.D/H.C.F can also be determined by using prime factorization

The Greatest Common Divisors (G.C.D) is also refered to as the Highest Common Factors (H.C.F) .The H.C.F of two natural numbers is the largest factor that they have in common.for example, the prime factorisation of 144 and 66 are:

144 = 2 x 2 x 2 x 2 x 3 x 3

66 = 2 x 3 x11

Only 2 and 3 are common to both factorisation and so the the highest common factor that these two number have in common (HCF) is 2 x3 = 6


To find G.C.D/H.C.F of a set of numbers using factorisation:

i). Express each number as a product of prime factor ii). Find the common prime factors iii). multiply the common prime factors together to give GCD/H.C.F


Find the G.C.D/H.C.F for the following set: { 60,90,135 }

60= 2x 2 x 3x 5

90= 2x3x 3x5

135= 3x3x 3 x5

Common prime factors are 3 and 5

G.C.D/H.C.F= 3 x 5 = 15

The concept of G.C.D/H.C.F is applicable in many real life situations:

Given two containers of capacity 20 litres and 30 litres respectively. Find the capacity of largest jug that can be used to fill each of containers exactly.

Solution  

20 = 2 x 2 x 5 ,

30 = 2 x 3 x 5

Common factors are 2 and 5

G.C.D = 2 X 5 = 10


Four ribbons of colours red , blue green and yellow of lengths 18m, 24m, 36m and 42 m respectively are to be used for wrapping a number of birthday gifts. The owner wishes to cut them into shorter pieces of equal length. find the greatest possible length if no ribbon is left over.

Solution

Write the numbers 18,24,36 and 42 as a product of their prime factors

18 = 2 x 3 x 3

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

42= 2 x 3 x 7

The common factors are 2 and 3

The G.C.D is 2 x 3 = 6 . Therefore the greatest possible length is 6m



3. Divisibility test for prime numbers, that is2,3,5,11

A number is divisible by:

2 - If the last digits is zero or even e.g 30,102, 6, 76

3 - If the sum of the digits are divisible by 3 ,

for example

1 2 3 .........{1+ 2 + 3 = 6}

34 5 ........ { 3 + 4 + 5 = 12}

2 0 4 .........{ 2 + 0 + 4 = 6}

5- If the last digit is 0 or 5 e.g. 50, 65, 70

11 - If the last sum of the alternate digit are equal

or their difference is a multiple of 11,

e.g

1 8 7 ...... ( 1 +7) - 8= 0

8 1 9 3 9....{8 + 9 + 9} -{1 +3} = 22

Hello

Have fun as you go through this lesson on Greatest Common Divisor or Highest Common Factor.

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