1445 = 1190*1 + 255
1190 = 255*4 + 170
255 = 170*1 + 85
170 = 85*2 + 0
So, now the remainder is 0, then HCF is 85
Now,
85 = 255 - 170
(1445 - 1190) - (1190 - 255*4)
⇒ 1445 - 1190 - 1190 + 255*4
⇒ 1445 - 1190*2 + (1445 - 1190)*4
⇒ 1445 - 1190*2 + 1445*4 - 1190*4
⇒ 1445*5 - 1190*6
⇒ 1190*(- 6) + 1445*5
1190m + 1445n , where m = - 6 and n = 5
Alternate solution
We know that Euclid's division Lemma is x and y for any two positive integers, there exist unique integers q and r satisfactorily x = yq + r, where 0 ≤ r <y. In case r=0 then y will be the HCF.
1445=1190x1+255
1190=255x4+170
255=170x1+85
170=85x2+0
We have found r=0
Hence, HCF(1190,1445)=85
So, now
85 = 255 - 170
=(1445-1190)-(1190-1020)
=(1445-1190)-(1190-255x4)
=1445-1190-1190+255x4
=1445-2×1190+(1445-1190)x4
=1445-2×1190+1445x4-1190x4
=1445+1445×4-2×1190-1190×4
=1445x5-1190x6
=-1190×6+1445×5
=1190x(-6)+1445x5
=1190m+1445n
(where m=-6 and n=5)
More clearly
Euclid's division lemma :
Let a and b be any two positive Integers.
Then there exists two unique whole numbers
q and r such that
a = bq + r ,
0 ≤ r < b
Alternatively ,
dividend = divisor × quotient + remainder
*******************************************
Now start with the larger interger , that is
1445 , apply the division lemma
1445 = 1190 × 1 + 255 ----( 1 )
1190 = 255 × 4 + 170 -----( 2 )
255 = 170 × 1 + 85 --------( 3 )
170 = 85 × 2 + 0 -----------( 4 )
The remainder has now become zero . so ,
our procedure stops .
Since the divisor at this stage is 85, the
HCF ( 1190 , 1445 ) = 85
Now ,
____
85 = 255 - 170 [ from ( 1 ) ]
= [ 1445 - 1190 × 1 ] - [ 1190 - 255 × 4 ]
[ from ( 1 ) and ( 2 ) ]
= 1445 - 1190 - 1190 + 255 × 4
= 1445 - 2 × 1190 + ( 1445 - 1190 × 1 ) × 4
[ from ( 1 ) ]
= 1445 - 2 × 1190 + 4 × 1445 - 4 × 1190
= 1190 ( - 6 ) + 1445 ( 5 )
compare with HCF = 1190m + 1445n
we easily conclude that ,
m = - 6 ;
n = 5 :
I hope this helps you.