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Question

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


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Solution

Step 1 : Find the HCF of given numbers

Given integers are81 and 237such that ,237>81

Applying division lemma to 81and 237, we get
237=81×2+75 …….(i)
Since the remainder is 750. So, apply the division lemma to the divisor 81 and the remainder 75 to get
81=75×1+6 ……(ii)
We consider the new divisor 75 and the new remainder 6 and apply division lemma, to get
75=6×12+3 ……(iii)

We consider the new divisor 6 and the new remainder 3 and apply division lemma, to get
6=3×2+0 ……(iv)
At this stage the remainder is zero. So, that last divisor or the non-zero remainder at the earlier stage i.e. 3is the HCF of 81 and 237.

Step 2 : Express the HCF in the form of a linear combination of the given numbers

Given numbers: 81 and 237.

Rewriting the equation (iii)
75=(6×12)+33=75(6×12)

By replacing 6 in the above expression, from equation (ii)
3=75(8175×1)×123=75(81×12)+(75×12)3=75(1+12)(81×12)3=75×1381×12

By replacing 75 from equation(i) as
3=237(81×2)×1381×123=(237×13)(81×26)(81×12)3=237×1381×383=237×13+81×-38
3=237x+81y, where x=13 and y=-38.

Hence, the HCF of81 and 237 is 3and the linear combination can be expressed as 3=237x+81y, where x=13 and y=-38.


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