Question

if the HCF of 65 and 117 is of the form (65m-117) then find the value of m

Answer 1

117=65*1+52
65=52*1+13
52=13*4+0
hcf of 117 and 65 is 13
65m-117=13
65m=130
m=130/65
m=2

Answer 2

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 117 > 65
We now try to express this as
117 = 65c + d for something and continue till d = 0
Now, 117 = 65*1 + 52

We see that, Remainder is not 0 .

So repeat the same with 52 , 65( 65 > 52 )
65 = 52 * 1 + 13

Still the remainder, is not zero.

Again repeating with ( 13 , 52 )
52 = 13 * 4 + 0

Therefore, The required HCF is 13

From the above equations,
13 = 65 - ( 52 )
13 = 65 - ( 117- 65)
13 = 65*2 - 117

Given that ,HCF ( 65 , 117 ) = 65m - 117

Now, 65 * 2 - 117 = 65m - 117

So, m = 2



Hope helped!