Question

use euclid's algorithm to find the HCF of 1190 and 1445. Express the HCF in the form 1190m+1445n.​

Answer 1

Answer:

m = - 6 and n = 5

Step-by-step explanation:

Solution :-

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 , Solution :-

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

Answer.

Answer 2

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