EXPRESS HCF OF TWO NUMBERS IN LINEAR COMBINATION

Problem 1 :

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

Solution :

237 > 81

237 = 81 x 2 + 75

81 = 75 x 1 + 6

75 = 6 x 12 + 3

6 = 3 x 2 + 0

HCF of (81 and 237) is 3.

Linear combination :

237 = 81x + 237y

Problem 2 :

Find the HCF of 65 and 117 and express it in the form

65m + 117n

Solution :

117 > 65

117 = 65 x 1 + 52

65 = 52 x 1 + 13

52 = 13 x 4 + 0

HCF of (117 and 65) is 13.

13 = 65(2) + 117(-1)

HCF = 65m + 117n

m = 2 and n = -1

Problem 3 :

If the HCF of 210 and 55 is expressible in the form of 

210x + 55y

Also show that x and y are not unique.

Solution :

First let us find HCF of 210 and 55.

210 > 55

210 = 55 ⋅ 3 + 45

55 = 45 ⋅ 1 + 10

45 = 10 ⋅ 4 + 5

10 = 5 ⋅ 2 + 0

HCF of 210 and 55 is 5.

HCF = 210x + 55y

5 = 210x + 55y

x = 1, then y = 

5 = 210(1) + 55y

5 - 210 = 55y

55y = -205

y = -205/55

Since we get decimal, we can apply some other value for x.

x = 5, then y =

5 = 210(5) + 55y

5 = 1050 + 55y

5 - 1050 = 55y

-1045 = 55y

y = -1045/55

y = -19

So, values of x and y are 5 and -19 respectively. By applying some more values for x, we will get different values for y.