Question

Se considera numerele naturale x,y,z care verifica relatia 2y-8z-7x=0
Demonstrati ca x³(2y-z) este divizibil cu 56.

Answer 1

2y-8z-7x=0
1) 2y-8z=7x
2(y-4z)=7x ⇒ x e par, x=2k
relatia  1) se mai scrie :
2y-z=7x+7y
2y-z=7(x+y) ⇒ 7|(2y-z) ⇒ (2y-7)=m7

x^3(2y-z) = 2^3 x k^3 x m x 7 = 8x7(mk^3) ⇒ 56 |[x^3(2y-z)] c.c.t.d

Answer 2

2y-8z-7x=0
2y-8z=7x
2y-(7z+z)=7x
2y-z=7x+7z
2y-z=7(x+z)
2y-z=M7
2y-z=7q
q∈N
2y-8z-7x=0
2y-8z=7x
2y=7x+8z
2y=6x+8z+x
2y=2(3x+4z)+x
2y=M2+x
M2=M2+x
x=M2
x=2k
k∈N
x^3(2y-z)=(2k)^3(7q)=(2^3k^3)*7q=8k^3*7q=(8*7)(k^3*q)=56*(k^3*q)
56Ix^3(2y-z)