Express the HCF of 468 and 222 as 468 x + 222 y where x, y are integer

1.	Express the HCF of 468 and 222 as 468 x + 222 y where x, y are integers in two different ways.
Answer» Given, we need to express the H.C.F. of 468 and 222 as 468 x + 222 y where x, y are integers in two different ways. So, here the integers are: 468 and 222, and 468 > 222 Then, by applying Euclid’s division lemma, we get 468 = 222 x 2 + 24……… (1) Since the remainder ≠ 0, so apply division lemma on divisor 222 and remainder 24 222 = 24 x 9 + 6………… (2) Since the remainder ≠ 0, so apply division lemma on divisor 24 and remainder 6 24 = 6 x 4 + 0……………. (3) We observe that remainder is 0. So, the last divisor 6 is the H.C.F. of 468 and 222 Now, in order to express the HCF as a linear combination of 468 and 222, we perform 6 = 222 – 24 x 9 [from (2)] = 222 – (468 – 222 x 2) x 9 [from (1)] = 222 – 468 x 9 + 222 x 18 6 = 222 x 19 – 468 x 9 = 468(-9) + 222(19) ∴ 6 = 468 x + 222 y, where x = -9 and y = 19.

Express the HCF of 468 and 222 as 468 x + 222 y where x, y are integers in two different ways.

Answer»

Given, we need to express the H.C.F. of 468 and 222 as 468 x + 222 y where x, y are integers in two different ways.

So, here the integers are: 468 and 222, and 468 > 222

Then, by applying Euclid’s division lemma, we get

468 = 222 x 2 + 24……… (1)

Since the remainder ≠ 0, so apply division lemma on divisor 222 and remainder 24

222 = 24 x 9 + 6………… (2)

Since the remainder ≠ 0, so apply division lemma on divisor 24 and remainder 6

24 = 6 x 4 + 0……………. (3)

We observe that remainder is 0.

So, the last divisor 6 is the H.C.F. of 468 and 222

Now, in order to express the HCF as a linear combination of 468 and 222, we perform

6 = 222 – 24 x 9 [from (2)]

= 222 – (468 – 222 x 2) x 9 [from (1)]

= 222 – 468 x 9 + 222 x 18

6 = 222 x 19 – 468 x 9

= 468(-9) + 222(19)

∴ 6 = 468 x + 222 y, where x = -9 and y = 19.