Question

find the HCF of 1190 and 1445 Express the HCF in the form 1190 M + 1445n
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Answer 1

Answer:

HCF(1190,1445)=85

Step-by-step explanation:

Euclid's division algorithm :

Let a and b are 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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Now ,

applying the division lemma to 1445 and

1190 ,

1445 = 1190 × 1 + 255 ----( 1 )

since the remainder is not equal to zero ,

we 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 has become zero , so

our procedure stops .

Since the divisor at this stage is 85 .

Therefore ,

HCF( 1445 , 1190 ) = 85.

Now ,

85 = 255 - 170 [ from ( 3 ) ]

= [ 1445 - 1190×1 ] - [ 1190 - 255 × 4 ]

[ from ( 1 ) and from ( 2 ) ]

= 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

compare this with ,

85 = 1190m + 1445n [ given ]

m = -6 ,

n = 5

I hope this helps you.