Fie n€N*. Determinați ultima cifră a numărului a=1•2•3•4•...•n+2004^n^2

Răspuns :


pt n, par,n² par, U(2004^2p)=6
pt n impar, n²impar U(2004^(2p+1))=4

pt n=1,U(a)=1+4=5
ptn=2, U(a)=2+6=8
ptn=3, U(a)=U(6+4)=0
ptn=4, U(a)=U(24+6)=0
ptn≥5, U(a) =U(0+ U(2004^n)=U(0+4 )=4, pt n =2p+1, p∈N
                                               =U(0+6)=6, pt n=2p, p∈N