Question

using euclids lemma find rh hcf of 92690,7378,7161

Answer 1

as there are there are three numbers first we will find the hcf of any two numbers then we will find the hcf of the other number with the hcf received
Set up a division problem where a is larger than b. 
a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.

92690 ÷ 7378 = 12 R 4154    (92690 = 12 × 7378 + 4154)

7378 ÷ 4154 = 1 R 3224    (7378 = 1 × 4154 + 3224)

4154 ÷ 3224 = 1 R 930    (4154 = 1 × 3224 + 930)

3224 ÷ 930 = 3 R 434    (3224 = 3 × 930 + 434)

930 ÷ 434 = 2 R 62    (930 = 2 × 434 + 62)

434 ÷ 62 = 7 R 0    (434 = 7 × 62 + 0)

When remainder R = 0, the HCF is the divisor, b, in the last equation. HCF = 62now as there are three numbers we should find the hcf of 7161 and the hcf we got just now that is 62
7161 ÷ 62 = 115 R 31    (7161 = 115 × 62 + 31)

62 ÷ 31 = 2 R 0    (62 = 2 × 31 + 0)

When remainder R = 0, the HCF is the divisor, b, in the last equation. HCF = 31