Question

multimea numerelor naturale care impartite la 5 dau restul 3 , iar impartite la 9 dau restul 7 .

Answer 1

folosim teorema impartirii cu rest
x:5=c,    rest 3⇒   x=5c+3        adunam primul si ultimul termen cu 2
x:9=c,    rest 7⇒   x=9c+7

x+2=5c+3+2      ⇒x+2=5c+5⇒     x+2=5(c+1)
x+2=9c+7+2      ⇒x+2=9c+9⇒     x+2=9(c+1)        ⇒

x+2∈M5UM9    ⇒x+2∈M[5,9]
[5,9]=45
x+2∈{0, 45, 90, 135, 180..........}
Daca x+2=45⇒  x=45-2⇒x=43
daca x+2=90⇒  x=88
daca x+2=135⇒ x=133
daca x+2=180⇒ x=178
...........................................
N={43, 88,133, 178.......}