Question

using division algorithm to find HCF of 1190 and 1445 Express the HCF in the form of 1190 M + 1445 n

Answer 1

Heya..!!!



Euclid's division algorithm :

Let a and b are two positive Integers .

We know that

q and r such that

a = bq + r ,

0 ≤ r < b


Now ,

applying the division lemma to 1445 and 1190 ,

1445 = 1190 × 1 + 255 -------(1)

∴ the remainder is not equal to zero ,

Now , apply the division lemma to 1190 and

255

1190 = 255 × 4 + 170 ---(2)

255 = 170 × 1 + 85 ------(3)

170 = 85 × 2 + 0----------(4)

The remainder zero ,


∵the divisor at this stage is 85 .

∴ HCF( 1445 , 1190 ) = 85.

Now ,

85 = 255 - 170---------from---(3)

= ( 1445 - 1190×1 ) - ( 1190 - 255 × 4 )------from--(1) & --(4)


= 1445 - 1190 - 1190 + 255 × 4

= 1445 - 2 × 1190 + ( 1445 - 1190 ) × 4---&-from---(1)

= 1445 - 2 × 1190 + 4 × 1445 - 4 × 1190

= 5 × 1445 - 6 × 11 90

85 = 1445 ( 5 ) + ( - 6 ) 1190

Now on comparing,

We get,


85 = 1190m + 1445n [ given ]

m = -6 ,

n = 5

I HOPE ITS HELP YOU