Question

Express HCF of 210 and 55 in the form of 210 x + 55 y​

Answer 1

Step-by-step explanation:

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

Answer 2

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$by \: euclid \: division \: lemma \\ 210 = 55 \times 3 + 45 \\ 55 = 45 \times 1 + 10 \\ 45 = 10 \times 4 + 5 \\ 10 = 5 \times 2 + 0 \\ \therefore \: hcf \: of \: 210 \: and \: 55 \: is \: 5 \\ now \\ 5 = 45 - (10 \times 4) \\ 5 = 45 - (55 - 45) \times 4 \\ 5 = 45 - (55 \times 4 - 45 \times 4) \\ 5 = 45 - 55(4) + 45(4) \\ 5 = 45 + 45(4) - 55(4) \\ 5 = 45(5) - 55(4) \\ 5 = (21 0 - 55 \times 3)5 - 55 \times 4 \\ 5 = 210(5) - 55(15) - 55(4) \\ 5 = 210(5) - 55(19) \\ \therefore \: x = 5 \: and \: y =-19$