If d is the hcf of …

We need to find the H.C.F. of 506 and 1155 and express it as a linear combination of 506 and 1155.
By applying Euclid’s division lemma
1155 = 506 {tex}\times{/tex} 2 + 143.
506 = 143 {tex}\times{/tex} 3 + 77.
143 = 77 {tex}\times{/tex} 1 + 66.
77 = 66 {tex}\times{/tex} 1 + 11.
66 = 11 {tex}\times{/tex} 6 + 0.
Therefore, H.C.F. = 11.
Now, 11 = 77 - 66 {tex}\times{/tex} 1 = 77 - [143 - 77 {tex}\times{/tex} 1] {tex}\times{/tex} 1           {∵ 143 = 77 {tex}\times{/tex} 1 + 66}
= 77 - 143 {tex}\times{/tex} 1 + 77 {tex}\times{/tex} 1
= 77 {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 1
= [506 - 143 {tex}\times{/tex} 3] {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 1  {∵ 506 = 143 {tex}\times{/tex} 3 + 77 }
= 506 {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 6 - 143 {tex}\times{/tex} 1
= 506 {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 7
= 506 {tex}\times{/tex} 2 - [1155 - 506 {tex}\times{/tex} 2] {tex}\times{/tex} 7 {∵1155 = 506 {tex}\times{/tex} 2 + 143 }
= 506 {tex}\times{/tex} 2 - 1155 {tex}\times{/tex} 7 + 506 {tex}\times{/tex} 14
= 506 {tex}\times{/tex} 16 - 1155 {tex}\times{/tex} 7
Hence obtained.