Question-172662

Question Number 172662 by SANOGO last updated on 29/Jun/22

Answered by Mathspace last updated on 29/Jun/22

a) Δ (x, y) = \frac{x y}{M (x, y)} = \frac{1512}{252} = 6

x = p d a n d y = q d w i t h (p, d) = 1

x + y = 1512 \Rightarrow d (p + q) = 1512

\Rightarrow p + q = \frac{1512}{6} = 252

\Rightarrow (p, q) \in {(1, 251), (2, 250), (3, 249) \dots (251, 1)}

w e e x t r a c t t h e c o u p l e s w i c h v e r i f y

Δ (p, q) = 1 \dots

\Rightarrow x = 6 p a n d y = 6 q \dots

Commented by SANOGO last updated on 30/Jun/22

t h a n k y o u

Commented by Mathspace last updated on 30/Jun/22

x + y = p d + q d = (p + q) d

= 6.252 = 1512

Answered by puissant last updated on 30/Jun/22

b)

p o s o n s δ = p g c d (x; y) d i r e c e l a r e n v o i a :

\exists (x^{'}; y^{'}) \in N^{2} / x = δ x^{'}; y = δ y^{'} e t x^{'} \land y^{'} = 1

x y = p g c d (x; y) ∙ p p c m (x; y)

\Rightarrow p p c m (x; y) = δ x^{'} y^{'} = 1440

e n m e m e t e m p s, δ (x^{'} + y^{'}) = 276

d o n c δ / 1440 e t δ / 276 \dots .

A t o i d e c o n c t i n u e r .

Answered by mr W last updated on 30/Jun/22

a)

x y = 1512 = 2^{3} \times 3^{3} \times 7

L C M = 252 = 2^{2} \times 3^{2} \times 7

n u m b e r 1 = 2 \times 3 \times 1

n u m b e r 2 = 2^{2} \times 3^{2} \times 7

(x, y) = (2 \times 3 \times 1, 2^{2} \times 3^{2} \times 7) = (6, 252) o r (252, 6)

(x, y) = (2^{2} \times 3 \times 1, 2 \times 3^{2} \times 7) = (12, 126) o r (126, 12)

(x, y) = (2 \times 3^{2} \times 1, 2^{2} \times 3 \times 7) = (18, 84) o r (84, 18)

(x, y) = (2 \times 3 \times 7, 2^{2} \times 3^{2} \times 1) = (42, 36) o r (36, 42)

Answered by Rasheed.Sindhi last updated on 30/Jun/22

(c) L e t x = 24 m & y = 24 n w h e r e

(m, n) = 1

x y = 24 m \cdot 24 n = 576 m n = 16128

m n = 28

m n = 1 \cdot 28 = 28 \cdot 1

m n = 4 \cdot 7 = 7 \cdot 4

(m, n) = (1, 28), (28, 1), (4, 7), (7, 4)

(x, y) = (24 m, 24 n)

= (24, 672), (672, 24), (96, 168), (168, 96)