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Find the HCF of 1190and 1445 …

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Find the HCF of 1190and 1445 in the. form of 1190m+1445n

Posted by Sachin Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Here we have to find HCF of 1190 and 1445 and express the HCF in the form 1190m + 1445n.
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

0Thank You

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