Find H. C. F at 592 …

We need to find the H.C.F. of 592 and 252 and express it as a linear combination of 592 and 252.

By applying Euclid’s division lemma

592 = 252 x 2 + 88

Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 88

252 = 88 x 2 + 76

Since remainder ≠ 0, apply division lemma on divisor 88 and remainder 76

88 = 76 x 1 + 12

Since remainder ≠ 0, apply division lemma on divisor 76 and remainder 12

76 = 12 x 6 + 4

Since remainder ≠ 0, apply division lemma on divisor 12 and remainder 4

12 = 4 x 3 + 0.

Therefore, H.C.F. = 4.

Now, 4 = 76 – 12 x 6

= 76 – 88 – 76 x 1 x 6

= 76 – 88 x 6 + 76 x 6

= 76 x 7 – 88 x 6

= 252 – 88 x 2 x 7 – 88 x 6

= 252 x 7- 88 x 14- 88 x 6

= 252 x 7- 88 x 20

= 252 x 7 – 592 – 252 x 2 x 20

= 252 x 7 – 592 x 20 + 252 x 40

= 252 x 47 – 592 x 20

= 252 x 47 + 592 x (-20)

Hence obtained.