Question

Use Euclid's algorithm to find HCF of 1190 and 1445 . Express the HCF in the form 1190m + 1445n .

Answer 1

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

Given:

Two numbers are 1190 and 1445.

To find:

HCF of 1190 and 1445 by using Euclid's algorithm and express the HCF in the form 1190m+1445n.

Solution:

By using Euclid's algorithm

$1445=1190\times 1+255$

$1190=255\times 4+170$

$255=170\times 1+85$

$170=85\times 2+0$

Hence, HCF of 1190 and 1445=85

$85=255-170$

$85=(1445-1190)-(1190-255\times 4)$

$85=1445-1190\times 2+255\times 4$

$85=1445-1190\times 2+(1445-1190)4$

$85=1445-1190\times 2+1445\times 4-1190\times 4$

$85=1445\times 5-1190\times 6$

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

$85=1445n+1190m$

Where m=-6,n=5