1.

Do the given problems in repeated subtraction method1. 56 and 122. 320,120 and 95

Answer»

1. 56 and 12

56 & 12

Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 = 8 = 4
m – n = 8 – 44. now m = n

HCF of 56 & 12 is 4

2. 320, 120 and 95

Let us take 320 & 120 first m = 320, n = 120

m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = w = 40 → HCF of 320, 120

Now let us find HCF of 40 & 95

m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5

HCF of 40 & 95 is 5 10 – 5 = 5

HCF of 320, 120 & 95 is 5


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