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Answer 1

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 2

∵ D( 1155, 506 ) .

→ 1155 = 506 × 2 +143.

→ 506 = 143 × 3 + 77.

→ 143 = 77 × 1 + 66.

→ 77 = 66 × 1 + 11.

→ 66 = 11 × 6 + 0.

∴ 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) .

∴ D = 1155(-7) + 506(16) .

x = -7 and y = 16.