# Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237

Hi!
Here is the proof to your query.

Among 81 and 237; 237 > 81
Since 237 > 81, we apply the division lemma to 237 and 81 to obtain
237 = 81 × 2 + 75  … Step 1
Since remainder 75 ≠ 0, we apply the division lemma to 81 and 75 to obtain
81 = 75 × 1 + 6  … Step 2
Since remainder 6 ≠ 0, we apply the division lemma to 75 and 6 to obtain
75 = 6 × 12 + 3  … Step 3
Since remainder 3 ≠ 0, we apply the division lemma to 6 and 3 to obtain
6 = 3 × 2 + 0     … Step 4
In this step the remainder is zero. Thus, the divisor i.e. 3 in this step is the H.C.F. of the given numbers
The H.C.F. of 237 and 81 is 3

From Step 3:
3 = 75 – 6 × 12    … Step 5
From Step 2:
6 = 81 – 75 × 1
Thus, from Step 5, it is obtained
3 = 75 – (81 – 75 × 1) × 12
⇒ 3 = 75 – (81× 12 – 75 × 12)
⇒ 3 = 75 × 13 – 81× 12     … Step 6
From Step 1;
75 = 237 – 81 × 2
Thus, from Step 6;
3 = (237 – 81 × 2) × 13 – 81× 12
⇒ 3 = (237 × 13 – 81 × 26) – 81× 12
⇒ 3 = 237 × 13 – 81 × 38
⇒ H.C.F. of 237 and 81 = 237 × 13 + 81 × (–38)

∴HCF, 3 can be expressed as linear combination of 81 and 237 as 3 = 81x + 237y, where x and y are not unique.

Cheers!

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can u plz tell me the chapter as soon as possible so that i may help u

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Its first chapter Real Numbers of class 10, i hope u r in 11th ryt?, i m also trying on this question

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no I am appearing for my 12th boards....

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please help me, i m in need of it!!, i hve to give exams

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yup i am tryin. Don't worry....

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By using euclid's division lemma

237=81 x 2 + 75

81=75 x 1 + 6

75= 6  x  12 + 3

6= 3  x  2 + 0

Therefore the H.C.F of 81 and 237 is 3.

but I am not getting the rest of the ques...

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are u still there?

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http://cbse.meritnation.com/discuss/question/792032

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i m not geeting sixth step...13?

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What are you looking for?