CBSE Class 10 Answered

Since 237 > 81, we apply the division lemma to 237 and 81 to obtain

237 = 81 x 2 + 75 … Step 1

Since remainder 75 ? 0, we apply the division lemma to 81 and 75 to obtain

81 = 75 x 1 + 6 … Step 2

Since remainder 6 ? 0, we apply the division lemma to 75 and 6 to obtain

75 = 6 x 12 + 3 … Step 3

Since remainder 3 ? 0, we apply the division lemma to 6 and 3 to obtain

6 = 3 x 2 + 0 … Step 4

In this step the remainder is zero. Thus, the divisor i.e. 3 in this step is the H.C.F. of the given numbers

The H.C.F. of 237 and 81 is 3

From Step 3:

3 = 75 – 6 x 12 … Step 5

From Step 2:

6 = 81 – 75 x 1

Thus, from Step 5, it is obtained

3 = 75 – (81 – 75 X 1) x 12

? 3 = 75 – (81x 12 – 75 x 12)

? 3 = 75 x 13 – 81x 12 … Step 6

From Step 1;

75 = 237 – 81 x 2

Thus, from Step 6;

3 = (237 – 81 x 2) x 13 – 81X 12

? 3 = (237 x 13 – 81 x 26) – 81x 12

? 3 = 237 x 13 – 81 x 38

? H.C.F. of 237 and 81 = 237 x 13 + 81 x (–38)

The above relationship is the representation of H.C.F. of 237 and 81 as linear combination of these two.