see the attached document Given
Answers
Question :-
→ D is the HCF of 1155 and 506. Find x and y satisfying D = 1155x + 506y also show x and y are not unique.
Answer :
→ x = -7 and y = 16.
Step-by-step explanation :
\because∵ D( 1155, 506 ) .
→ 1155 = 506 × 2 +143.
→ 506 = 143 × 3 + 77.
→ 143 = 77 × 1 + 66.
→ 77 = 66 × 1 + 11.
→ 66 = 11 × 6 + 0.
\therefore∴ D = 11 .
= 77 - 66 × 1 .
= 77 - [ 143 - 77 ] .
= 77 - 143 + 77 .
= 77 × 2 - [ 1155 - 506 × 2 ] .
= [ 506 - 143 × 3 ] (2) - [ 1155 - 506 × 2 ] .
= [ 506 - ( 1155 - 506 × 2 ) × 3 ] (2) - [ 1155 - 506 × 2 ] .
= 506(2) - [ 1155 - 506(2) ](6) - 1155 + 506(2) .
= 506(2) - 1155(6) + 506(12) - 1155(1) + 506(2) .
= 506(16) - 1155(7) .
\therefore∴ D = 1155(-7) + 506(16) .
And, it given,
→ D = 1155x + 506y.
→ 1155x + 506y = 1155(-7) + 506(16) .
By comparing, we get,
\therefore∴ x = -7 and y = 16.
Hence, it is solved.
Answer:
D is the hcf of 1155 and 506.
D= 1155x + 506x
D= 11
x= -7 and y = 16