Express the HCF of 468 and 222 as 468x +22y where x, y are integers in

1.	Express the HCF of 468 and 222 as 468x +22y where x, y are integers in twodifferent ways.
Answer» Given integers are 468 and 222 where 468 > 222. By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24 …(i) Since remainder ≠ 0, apply division lemma on division 222 and remainder 24 222 = 24 × 9 + 6 …(ii) Since remainder ≠ 0, apply division lemma on division 24 and remainder 6 24 = 6 × 4 + 0 …(iii) We observe that the remainder = 0, so the last divisor 6 is the HCF of the 468 and 222 From (ii) we have 6 = 222 – 24 × 9 6 = 222 – [468 – 222 × 2] × 9 [Substituting 24 = 468 – 222 × 2 from (i)] 6 = 222 – 468 × 9 – 222 × 18 6 = 222y + 468x, where x = −9 and y = 19 6 = 222 × 19 – 468 × 9 Like if you find it useful

Express the HCF of 468 and 222 as 468x +22y where x, y are integers in twodifferent ways.

Answer»

Given integers are 468 and 222 where 468 > 222.

By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24 …(i) Since remainder ≠ 0,

apply division lemma on division 222 and remainder 24

222 = 24 × 9 + 6 …(ii) Since remainder ≠ 0,

apply division lemma on division 24 and remainder 6 24 = 6 × 4 + 0 …(iii)

We observe that the remainder = 0,

so the last divisor 6 is the HCF of the 468 and 222

From (ii) we have 6 = 222 – 24 × 9

6 = 222 – [468 – 222 × 2] × 9

[Substituting 24 = 468 – 222 × 2 from (i)]

6 = 222 – 468 × 9 – 222 × 18 6 = 222y + 468x, where x = −9 and y = 19

6 = 222 × 19 – 468 × 9

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