HIGHEST COMMON FACTORS (H.C.F)

After listing all the common factors of a given number, the highest among these numbers is the highest common factor.

Example

        1.    Find the highest common factor of 63, 36 and 27.

Solution

63 – 1, 3, 7, 9, 21, 63

36 – 1, 2, 3, 4, 6, 9, 12, 18, 36

27 – 1, 3, 9, 27

Common factors – 1, 3, 9

Highest common factor = 9.

Listing factors when finding the H.C.F of numbers is not easy at all, so using the above method to find H.C.F is not recommended. An easier method (index form method) of solving H.C.F is shown below.

Index form method

Step 1: Express each number as a product of prime factors in index form

63 = 32 x 7

36 = 22 x 32

27 = 33

Step 2: look for the base that is common to the three numbers.

Here, only base 3 is common to the three numbers. i.e. 32, 32 and 33 in the first, second and third number respectively.

Step 3: For H.C.F pick the base with the smallest power (32), we have two of them, just pick one.

Therefore, H.C.F = 32

= 9

A simpler and shorter (Table) method is also shown below.

Table method:

Step 1: We create a table and put each number in a column of the table, the first column of the table is left empty for the divisor.

  63 36 27

Step 2: We look for a number that can divide the three numbers without remainder. We start with 2 first, since 2 cannot divide the three numbers, then we try 3. 3 can divide the three numbers to give 21, 12 and 9, then we write 3 in the first column meant for the divisor and write out the various answers after the division under each number.

3 63 36 27
  21 12 9

Looking at the three numbers 21, 12 and 9; three can still divide them to give 7, 4 and 3.

3 63 36 27
3 21 12 9
  7 4 3

The three numbers 7, 4 and 3 above cannot be divided by any other number, so we conclude that

H.C.F = 3 x 3

= 9.

The subsequent examples will be solved using the index and table method.

Note: when looking for what can divide, we start with 2, then 3, 5, 7, 11... (prime numbers)

         2.    Find the H.C.F of 90, 54 and 144.

Solution

Index method

90 = 2 x 32 x 5

54 = 2 x 33

144 = 24 x 32

>> Common bases 2 and 3

>> Select the base with the smallest power 2 x 32

= 2 x 9

= 18.

Note: 2 also means 21.

Table method

  90 54 144

Always check whether 2 can divide the three numbers before testing for 3.

** 2 will divide all to give 45, 27 and 72

2 90 54 144
  45 27 72

** 2 cannot divide the three numbers (because of the 27 present), then we check 3.

** 3 will divide the three numbers to give 15, 9 and 24

2 90 54 144
3 45 27 72
  15 9 24

** 3 can still divide the three numbers to give 5, 3, and 8 respectively.

2 90 54 144
3 45 27 72
3 15 9 24
  5 3 8

** No other number can divide the above three numbers, so we calculate our H.C.F by multiplying the numbers on the first column.

H.C.F = 2 x 3 x 3

= 18.

        3.    Find the H.C.F of 180, 220 and 408

Solution

Index method

Express each as a product of prime factors in index form.

180 = 22 x 32 x 51

220 = 22 x 51 x 11

408 = 23 x 3 x 17

H.C.F = 22

= 4

Table method

2 180 220 408
2 90 110 204
  45 55 102

Looking at the three numbers 45, 55 and 102 above, no number (apart from 1) can divide all without remainder.

** 2 will only divide 102

** 3 will divide 45 and 102 only

** 5 will divide only 45 and 55

** 7 will not divide any of the three

** 11 will only divide 55.

Therefore, HCF = 2 x 2

= 4.

        4.    Find the H.C.F of 270, 585, 405 and 900

Solution

Index method

270 = 2 x 33 x 5

585 = 32 x 5 x 13

405 = 34 x 5

900 = 22 x 32 x 52

H.C.F = 32 x 5

= 45

Table Method

3 270 585 405 900
3 90 195 135 300
5 30 65 45 100
  6 13 9 20

H.C.F = 3 x 3 x 5
H.C.F = 45

Exercise E  

Find the H.C.F of the following

1. 18, 96 and 24

2. 12, 15, 24 and 9

3. 48, 84 and 132

4. 42, 30, 54 and 108

5. 288, 180 and 108

6. 144, 252 and 432

7. 75, 50, 125 and 150

8. 135, 360 and 225

9. 432, 528 and 384

10. 110, 264, 176 and 198

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