Question

the hcf of 79 and 97 is expressible in the form 97n-79m then value of m-n is

Answer 1

Hey there,
In Euclid division lemma, We try to express one number as a multiple of number and we continue till we get a smallest number which leaves 0 as remainder.

Given, Two numbers 79 , 97 .
First we determine which of them is the biggest.
So 97 > 79 .
We now try to express this as
97 = 79c + d for something and continue till d = 0
Now, 97 = 79 * 1 + 18 .

We see that, Remainder is not 0 .

So repeat the same with 18 , 79 ( 79 > 18 )
79 = 18*4 + 7

Still the remainder, is not zero.

Again repeating with ( 7 , 18 )
18 = 7 * 2 + 4
Now, 7 = 4 * 1 + 3
Also, 4 = 3 * 1 + 1
Now. 3 = 1 * 3 + 0

Therefore, The required HCF is 1 ,

From the above equations,
1 = 4 - 3 * 1
1 = 4 - ( 7 - 4 )
1 = 4 - 7 + 4
1 = 4 * 2 - 7
1 = ( 18 - 7 * 2 )*2 - 7
1 = 18 * 2 - 7 * 4 - 7
1 = 18 * 2 - 7*5
1 = 18 * 2 - ( 79 - 18*4 )
1 = 18*2 - ( 79 ) + 18*4
1 = 18*6 - 79
1 = (97 - 79 )* 6 - 79
1 = 97 * 6 - 79* 6 - 79
1 = 97 * 6 - 79 ( 6 + 1 )
1 = 97*6 - 79* 7

Given that ,HCF ( 79 , 97 ) = 97n - 79m

Now, 97n - 79m = 97*6-79*7

So, n = 6 and m = 7 .

Now. m-n =7 - 6 = 1 .

Hope helped!

Answer 2

Hi friend, I hope this helps you