Page 36 - SM Class 5 Inner.cdr

4. We will continue this process till we get remainder zero and divisor obtained
in end is the required H.C.F.

Let's take few examples for this :

2 4 8 4 9 2 1 Example 9 : Find the H.C.F. of 248 and 492.
To find the solution we will follow the following
2 4 8
2 4 4 2 4 8 1 method i.e. we divide bigger number 492 by smaller
2 4 4 one i.e. 248
4 2 4 4 61 So the divisor in the end is 4 so the H.C.F of the
2 4 4 given numbers is 4.
× 0

396 4 2 0 1
Example 10: Find the H.C.F. of 420 and 396.
3 9 6 To find the solution we will follow the following
2 4 3 9 6 16
method i.e. we divide bigger number 420 by
3 8 4
smaller one i.e. 396
1 2 3 8 4 32
So the divisor in the end is 12 so the H.C.F of
3 8 4
the given numbers is 12.
× 0

3. HCF by common division method

Step involved in common division method :

1. Arrange the given number in a row.
2. Divide the numbers by the lowest common prime number and write the
quotient below the numbers.
3. Further, divide the quotients by the lowest common prime number.

4. Repeat the process till there are no common prime numbers to divide by.
5. Find the product of the divisors. It is the required HCF.

Example 11 : Find the HCF of 48, 60, 72.

Solution :
2 48, 60, 72
2 24, 30, 36

3 12, 15, 18
4, 5, 6

So, HCF = 2 × 2 × 3 = 12

E
xercise 3.4

1. Find the HCF of the following numbers :

a. 18, 40 b. 30, 42 c. 60, 100 and 120 d. 18, 54 and 81

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Mathematics In Focus - 5