x= nr cautat
x>15
x:15=c1 , r=5, deci x=15c1+5
x:6=c2, r=5, deci x=6c2+5
c1 si c2 sunt numere intregi
15c1+5=6c2+5
prin urmare 15c1=6c2. Impartim la 3 si obtinem
5c1=2c2,
Dam valori c1=1 obtinem c2=5/2 care nu convine
c1=2 obtinem c2=5.
Prin urmare x=15*2+5=35