Use Euclid's algorithm to find HCF of 1190 and 1445. Express the HCF
in the form 1190m + 1445n.
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Answer:
HCF=85
85=1190*(-6) +1445*5
Comparing with 85 =1190m+1445n,
We get m= - 6 and n=5
See the detailed explanation of the solution below
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Step-by-step explanation:
Using Euclid's algorithm
a= bq + r
1445= 1190*1 + 255. (i)
1190= 255*4 + 170. (ii)
255= 170*1 + 85. (iii)
170= 85*2+0. (iv)
Therefore hcf =85
85= 255*1 - 170 from (iii)
85=(1445-1190) - [1190 - (255)*4]. From (i) and (ii)
85=(1445-1190) - [1190 - (1445-1190)*4] from (i)
85=(1445-1190)-1190+(1445-1190)*4
85= 1445-1190-1190+1445*4-1190*4
85=1445 - (2*1190) - (4*1190)+1445*4
85=1445+1445*4 - (6*1190)
85=1445*5-1190*6
85=1445*5+1190*(-6)
85=1190*-6+1445*5
Now equating this equation to 85=1190m+1445n
We get m= - 6 and n=5
Please mark my answer as the brainlliest :)
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