Question

Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237 i.e HCF (81,237)=81x+237y for some x and y.
for 20 points and brilliant in a paper​

Answer 1

Answer:

Step-by-step explanation:

237 > 81

BY APPLY DIVISION,

237 = 81 * 2 + 75                       (STEP-1)

REMAINDER IS 75

81 = 75 * 1 + 6                          (STEP-2)

REMAINDER IS 6

75 = 6 * 12 + 3                       (STEP-3)

REMAINDER IS 3

6 = 3 * 2 + 0                            ( STEP-4)

REMAINDER IS 0 SO,

H.C.F OF 81 AND 237 IS 3

Answer 2

Answer:

Given integers are 81 and 237 such that 81<237.

Applying division lemma to 81 and 237, we get

237=81×2+75 

Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get

81=75×1+6  

We consider the new divisor 75 and the new remainder 6 and apply division lemma to get

75=6×12+3  

We consider the new divisor 6 and the new remainder 3 and apply division lemma to get

6=3×2+0  

The remainder at this stage is zero. So, the divisor at this stage or the remainder at the earlier stage i.e. 3 is the HCF of 81 and 237.

To represent the HCF as a linear combination of the given two numbers, we start from the last but one step and successively eliminate the previous remainder as follows: 

From (iii), we have

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

⇒3=237x+81y, where x=13 and y=−38.

Now the HCF (say d) of two positive integers