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

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

Grade:10

3 Answers

Arun
25763 Points
4 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
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the attached solution to your question.
 
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
one year 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

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