If d is HCF to 40 and 65, find the value of integers x and y which satisfy d=40 x+65 y.

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Solution

HCF (40.65) is

$65 = 1 \times 40 + 25 . . . (1)$
$40 = 1 \times 25 + 15 . . . (2)$
$25 = 1 \times 15 + 10 . . . (3)$
$15 = 1 \times 10 + 5 . . . (4)$
$10 = 2 \times 5 + 0 . . . (5)$

$∴ H C F (40, 65) = 5$

$(4) \Rightarrow 5 = 15 - 1 \times 10$
$(3) \Rightarrow 5 = 15 - 1 \times (25 - 1 \times 15)$ = 15 - 25 + 15
= 2 \times 15 - 25

$(2) \Rightarrow 2 \times (40 - 1 \times 25) - 25 = 2 \times 40 - 2 \times 25 - 1 \times 25$
$= 2 \times 40 - 3 \times 25 - 1 \times 25$
$= 2 \times 40 - 3 \times 25$

(1) $\Rightarrow 5 = 2 \times 40 - 3 \times (65 - 1 \times 40)$
$= 2 \times 40 - 3 \times 65 + 3 \times 40$
$= 5 \times 40 - 3 \times 65$
$= 5 \times 40 + (- 3) 65$
=40x=65y

$∴ x = 5, y = - 3$

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