Using division algorithm, we may express HCF of two numbers as linear combination.
Step 1 :
In the given numbers, choose the larger number one.
Step 2 :
Decompose the larger one by the multiple of smaller one.
Step 3 :
Step 2 will have the form,
Divided = Divisor x quotient + Remainder
Repeat the process until we receive remainder as 0.
At last by dividing which value, we get the remainder as 0, that can be considered as HCF of the given numbers.
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.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM