U (1)=1
U(2!)=U(2)=2
U(3!)=U(6)=6
U (4!)=U(124)=4
U (5!)=U(620)=0
U (6!)=U(3720)=0
presupunem ca pt∀n≥5, U(n!)=0
verificata pt n=5
verificam pt n+1
U ((n+1)!)= U (n!* (n+1) =U (0* (n+1))= U(0)*U (n+1)= 0* U (n+1)=0, orice numar inmultit cu 0 da 0
Pn⇒P(n+1) deci relatia este demonstrata prin inductie completa
∀n≥5, U(n!)=0
Atunci U ( 5!) +U (6!) +...+ U ( 2016!)=U ( 0+0+...+0)=U(0)=0
si U ( A)= U (1+2+6+4+0)= U (13)=3, cerinta
