Page 63 - ICSE Math 6

Example 24: Find the HCF of 240 and 340.
Solution: As 240 < 340, therefore divide 340 by 240.
1
240 340
–240 2
100 240 (Dividing the previous divisor by the remainder)
–200 2
40 100
–80 2
20 40
–40
×
As 20 is the last divisor, therefore HCF of 240 and 340 is 20.
Division method for finding HCF of more than two numbers

Step 1: Find the HCF of any of the two given numbers.
Step 2: Find the HCF of the third number and the HCF obtained in step 1. Continue in this manner
to get the required HCF.

Example 25: Find the HCF of 250, 930 and 445.
Solution: First we find the HCF of 250 and 930.
3
250 930
–750 1
180 250 (Dividing the previous divisor by the remainder)
–180 2
70 180
–140 1
40 70
–40 1
30 40
–30 3
10 30
–30
×
Thus, the HCF of 250 and 930 is 10.
Now, we find the HCF of 10 and 445.
44
10 445
–40
45
–40 2
5 10
–10
×
Thus, the HCF of 10 and 445 is 5.
Hence, the HCF of 250, 930 and 445 is 5.

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