Arun
Last Activity: 7 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)