n = 2003a + r r ∈ {0,1,2,3,4.....2002} (2003 variante de impartire)
a ∈ {0, 1,2,3.....2002} ( 2003 valori )
total: 2003·2003 = 2003² impartiri posibile
de exemplu : ptr. r = n 0 ≤ n ≤ 2002
S = (2003·0 +n) + (2003·1 + n) + (2003·2+ n) + .............2003·(n+1) +n
Stotala = 2003[1+2+3+.....(n+1)]·2003² + S resturi
S resturi = 2003·0 + 2003 ·1 + 2003·2 + ..........+2003·2002 =
= 2003( 0 + 1 +2 +........+2002) = 2003·2002·2003/2 = 1001·2003² ⇒
⇒ S total divizibil cu 2003
