If d is the hcf of ( 1155 and 506) then find x,y satisfing d= 1155x +

1.	If d is the hcf of ( 1155 and 506) then find x,y satisfing d= 1155x + 506y
Answer» 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 lemma1155 = 506 {tex}\\times{/tex}\xa02 + 143.506 = 143 {tex}\\times{/tex}\xa03 + 77.143 = 77 {tex}\\times{/tex}\xa01 + 66.77 = 66 {tex}\\times{/tex}\xa01 + 11.66 = 11 {tex}\\times{/tex}\xa06 + 0.Therefore, H.C.F. = 11.Now, 11 = 77 - 66 {tex}\\times{/tex}\xa01 = 77 - [143 - 77 {tex}\\times{/tex}\xa01] {tex}\\times{/tex}\xa01 {∵ 143 = 77 {tex}\\times{/tex}\xa01 + 66}= 77 - 143 {tex}\\times{/tex}\xa01 + 77 {tex}\\times{/tex}\xa01= 77 {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa01= [506 - 143 {tex}\\times{/tex}\xa03] {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa01 {∵\xa0506 = 143 {tex}\\times{/tex}\xa03 + 77 }= 506 {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa06 - 143 {tex}\\times{/tex}\xa01= 506 {tex}\\times{/tex}\xa02 - 143 {tex}\\times{/tex}\xa07= 506 {tex}\\times{/tex}\xa02 - [1155 - 506 {tex}\\times{/tex}\xa02] {tex}\\times{/tex}\xa07 {∵1155 = 506 {tex}\\times{/tex}\xa02 + 143\xa0}= 506 {tex}\\times{/tex}\xa02 - 1155 {tex}\\times{/tex}\xa07 + 506 {tex}\\times{/tex}\xa014= 506 {tex}\\times{/tex}\xa016 - 1155 {tex}\\times{/tex}\xa07Hence obtained.

If d is the hcf of ( 1155 and 506) then find x,y satisfing d= 1155x + 506y