1)[tex](6x+5)\vdots 30=>6x+5=30k=>x= \frac{30k-5}{6} [/tex]
Deoarece 30k-5 se termina in 5 si nu exista multiplu de 6 care sa se termine in 5 , deducem ca nu exista un numar natural x asfel incat [tex](6x+5)\vdots 30[/tex]
2)[tex](2x+1)\vdots35=>2x+1=35k=>x= \frac{35k-1}{2} = 17k+ \frac{k-1}{2}=>\\
k=1,3,5,7,9,11,......\\
x=17,52,87,......[/tex]
