Find hcf of 210 and 55 and express as a linear combination

210 = 55* 3 + 45   ….....(i)

55 = 45 * 1 +10  ….........(ii)

45 = 10 *4 +5  …...........(iii)

10 = 5 *2 + 0

 

now from (iii), we get

45 = 10 *4 + 5

so,

5 = 45 – 10*4

5 = 45 – (55 – 45)*4

5 = 45 – 55*4 + 45*4

5 = 45 *5 – 55*4

5 = (210 – 55*3) *5 – 55*4

5 = 210*5 – 55*15 – 55*4

 

Hope it helps.

Thanks and regards,

Kushagra

Firstly we will find the hcf of 55 and 210 by applying Euclid's division algorithm
210>55
*210=55x3+45........(1)
*55=45x1+10.......(2)
*45=10x4+5..........(3)
*10=5x2+0..........(4)
so, HCF is 5
from equation 3 we have
5=45-(10x4)

5 = 45 – (55 – 45)*4. ( from equation 2)

5 = 45 – 55*4 + 45*4

5 = 45 *5 – 55*4

5 = (210 – 55*3) *5 – 55*4

5 = 210*5 – 55*15 – 55*4

5 = 210*5 – 55*19

5 = 210a + 55b

where a = 5

b=-19

thanks