x=7c+6⇒ x+1=7c+6+1⇒x+1=7c+7
x=6c+5⇒ x+1=6c+5+1⇒x+1=6c+6
x+1∈M[7,6] ⇒ [7,6]=42 M[42]={42,84,126,168,210....}
x+1=M[7,6]
x=M[7,6] -1
x=126-1 ⇒x=125
verificare
125:7=17 rest 6
125:6=20 rest 5
125 este div.cu5
cel mai mic nr.natural care satisface conditiile de mai sus este 125
