folosim teorema impartirii cu rest:D:I=C si rest r, D=I·C+r
x:3=c, rest 2 ⇒x=3·c+2 scadem primul si al doilea termen cu 2⇒
x:5=c, rest 2 ⇒x=5·c+2
x:4=c, rest 2 ⇒x=4·c+2
x-2=3c+2-2 ⇒x-2=3c
x-2=5c+2-2 ⇒x-2=5c ⇒ ⇒
x-2=4c+2-2 ⇒x-2=4c
x-2∈M[3,4,5]
[3,4,5]=60
M(60)={0, 60, 120, 180, 240.....}
daca x-2=60⇒x=62
daca x-2=120⇒x=162
daca x-2=180⇒x=182
daca x=240⇒x=242
nr.naturale cuprinse intre 100 si 200 sunt162 si 182
deci 162 si 182 impartite pe rand la 3, 4,5 dau toate restul 2