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# find  how of 81 and 237 and express it as linear combination of 81 and 237 i.e 237x =31x+237y for some  x and y

## 1 Answers

3 years ago

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.

Regards

Arun (askIITians forum expert)

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