Q.

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


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a

3, 94×237+(-275)×81

b

3, 94×237+(+275)×81

c

3, 95×238+(-275)×81

d

3, 94×237+(-277)×82 

answer is A.

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Detailed Solution

Given Integer = 81 ,237 Here,
81<237
Now, applying division lemma,
237=81×2+75…………….(1)
So we can see the remainder 750 Again from division lemma,
81=75×1+6…………………(2)
We can see the new remainder 60 
And the new divisor= 75,
75=6×12+3…………………(3)
When we take new divisor = 6 So, new remainder =3
6=3×2+0
Now we can see the remainder at this stage is 0.
Here, for the representation of HCF as the linear combination of the provided numbers, take equation (3) as,
3=75-6×12 
3=75-81-75×1×12 
3=75-12×81+12×75 
3=13×75-12×81 
3=13×237-81×2-12×81 
3=13×237-26×81-12×81 
3=13×237-38×81 
Suppose, x=12 and y=-38 so we get the linear equation as,
3=237x+81y 3=13×237-38×81+237×81-237×81 
3=13×237+237×81+-38×81-237×81
3=13+81×237+-38-237×81
 3=94×237-275×81 
3=94×237+(-275)×81 
Here, 3 is the HCF of given number which can be expressed as 94×237+(-275)×81.
Correct option is 1.
 
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