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# Express the Hcf of 468 and 222 as 468x + 222y Where X, Y Are Integers in Two Different Ways. - CBSE Class 10 - Mathematics

#### Question

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

#### Solution

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 = 222 × 19 – 468 × 9

⇒ 6 = 222y + 468x, where x = −9 and y = 19

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#### APPEARS IN

RD Sharma Solution for 10 Mathematics (2018 to Current)
Chapter 1: Real Numbers
Ex. 1.20 | Q: 22 | Page no. 28

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Solution Express the Hcf of 468 and 222 as 468x + 222y Where X, Y Are Integers in Two Different Ways. Concept: Euclid’s Division Lemma.
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