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

Arun
25750 Points
6 years ago
HCF of 210 & 55
210 = 55* 3 + 45   ….....(i)
55 = 45 * 1 +10  ….........(ii)
45 = 10 *4 +5  …...........(iii)
10 = 5 *2 + 0
hence HCF of 210 & 55 = 5

now from (iii), we get
4 5 = 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
5 = 210*5 – 55*19
5 = 210 x + 55 y
where x = 5, y = –19

hope it helps
4 years ago
Dear student,

HCF of 210 & 55
210 = 55* 3 + 45   ….....(i)
55 = 45 * 1 +10  ….........(ii)
45 = 10 *4 +5  …...........(iii)
10 = 5 *2 + 0
Hence HCF of 210 & 55 = 5

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
5 = 210*5 – 55*19
5 = 210 x + 55 y
where x = 5, y = –19

Hope it helps.
Thanks and regards,
Kushagra
RISHIKA
550 Points
4 years ago
HI FRIENDS
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