Find the HCF of the pair of integers and express it as a linear combin

1.	Find the HCF of the pair of integers and express it as a linear combination of them 592 and 252.
Answer» By applying Euclid’s division lemma on 592 and 252, we get 592 = 252 x 2 + 88……… (1) As the remainder ≠ 0, apply division lemma on divisor 252 and remainder 88 252 = 88 x 2 + 76………. (2) As the remainder ≠ 0, apply division lemma on divisor 88 and remainder 76 88 = 76 x 1 + 12………… (3) As the remainder ≠ 0, apply division lemma on divisor 76 and remainder 12 76 = 12 x 6 + 4………….. (4) Since the remainder ≠ 0, apply division lemma on divisor 12 and remainder 4 12 = 4 x 3 + 0……………. (5) Thus, we can conclude the H.C.F. = 4. Now, in order to express the found HCF as a linear combination of 592 and 252, we perform 4 = 76 – 12 x 6 [from (4)] = 76 – [88 – 76 x 1] x 6 [from (3)] = 76 – 88 x 6 + 76 x 6 = 76 x 7 – 88 x 6 = [252 – 88 x 2] x 7 – 88 x 6 [from (2)] = 252 x 7- 88 x 14 - 88 x 6 = 252 x 7- 88 x 20 = 252 x 7 – [592 – 252 x 2] x 20 [from (1)] = 252 x 7 – 592 x 20 + 252 x 40 = 252 x 47 – 592 x 20 = 252 x 47 + 592 x (-20)

Find the HCF of the pair of integers and express it as a linear combination of them 592 and 252.

Answer»

By applying Euclid’s division lemma on 592 and 252, we get

592 = 252 x 2 + 88……… (1)

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

252 = 88 x 2 + 76………. (2)

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

88 = 76 x 1 + 12………… (3)

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

76 = 12 x 6 + 4………….. (4)

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

12 = 4 x 3 + 0……………. (5)

Thus, we can conclude the H.C.F. = 4.

Now, in order to express the found HCF as a linear combination of 592 and 252, we perform

4 = 76 – 12 x 6 [from (4)]

= 76 – [88 – 76 x 1] x 6 [from (3)]

= 76 – 88 x 6 + 76 x 6

= 76 x 7 – 88 x 6

= [252 – 88 x 2] x 7 – 88 x 6 [from (2)]

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

= 252 x 7- 88 x 20

= 252 x 7 – [592 – 252 x 2] x 20 [from (1)]

= 252 x 7 – 592 x 20 + 252 x 40

= 252 x 47 – 592 x 20

= 252 x 47 + 592 x (-20)