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