Question

Express hcf of 48 and 18 as 48x + 18y and show the values of x and y are not unique

Answer 1

HCF of 48 and 18 :
factors of 18 = 2 × 3 × 3
factors of 48 = 2 × 2 × 2 × 2 × 3
We know, HCF is the highest common factors of given numbers
Hence, HCF{48, 18 } = common factors{48,18} = 2 × 3 = 6

Now, question said HCF {48, 18} is in the form of 48x + 18y
48x + 18y = 6
⇒8x + 3y = 1, there are many solution will be possible because we have two variable but given one equation . Hence values of x and y are not unique,
if x = 2 and y = -5 ⇒8 × 2 + 3 × (-5) = 1
Hence, x = 2 and y = -5
If x = 5 and y = -13 ⇒40 - 39 = 1
Hence, solution also will be x = 5 and y = -13 so, many solution will be possible.

Answer 2

Hi ,

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Euclid's division lemma :

Let a and b be two positive Integers .

Then there exist two unique whole numbers

q and r such that

a = bq + r ,

0 ≤ r < b

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Start with larger number ,that is 48

48 = 18 × 2 + 6

18 = 6 × 3 + 0

The remainder has now became zero , so

our procedure stops .

since the divisor at this stage is 6 ,

the HCF ( 48 , 18 ) = 6

according to the problem given ,

6 = 48x + 18y

This is a linear equation in two variables x

and y .

As we know that every linear equation in

two variables has infinitely many solutions.

example :

1 ) if x = - 1 , y = 3

6 = -48 + 54

2 ) if x = 2 , y = -5

6 = 48 ( 2 ) + 18 × ( -5 )

6 = 96 - 90

I hope this helps you.

: )