Question

if the hcf of81 and 273 can be written as 237m + 81n then the values of m and n are (a)m=13,n=-38 (b)m=13,n=38 (c)m=-13,n=38 (d)m=-13,n=-38​

Answer 1

Question

If the hcf of81 and 273 can be written as 237m + 81n then the values of m and n are (a)m=13,n=-38 (b)m=13,n=38 (c)m=-13,n=38 (d)m=-13,n=-38

Answer

Hcf (81,237) = 3

3= 237m+81n

1= 79m + 27n

m= 13

n=-38

Answer 2

Answer:

(a.) m=13 , n=(-38)

Step-by-step explanation:

❈ Given:-

   81<237 and these integers can be written as 237m+81n.

❈ To Find:-

    The values of m and n.

❈ Solution:-

We will apply Euclid Division Lemma to find the HCF of 81 and 273.

⇒ 237=81×2+75

⇒ 81=75×1+6

⇒ 75=6×12+3

⇒ 6=3×2+0

∴ The HCF of 81 and 273 is 3.

Now, we will start from the last step but one step and successively eliminate the previous remainder:

⇒ 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=237m+81n, where m=13 and n=(−38).

∴ The values of m and n will be 13 and (-38) respectively.