Question

sa se arate ca suma tuturor nr. nat. care dau aceleasi cat la imp. cu 2005 este divixibila cu 2005

Answer 1

   
Luam in considerare toate numerele n care impartite la 2005
care dau catul k si restul r.
unde:
r ∈ {0; 1; 2; 3; .......; 2004}
n, k ∈ N
Multimea acestor numere este:
D ={n | n : 2005 = k si rest r} = {2005k+0; 2005k+1; 2005k+2; ...; 2005k+2004}

Suma acestor numere este:

[tex]\displaystyle\\ 2005k+0+2005k+1+ 2005k+2+ \cdots +2005k+2014=\\\\ \underbrace{2005k+2005k+2005k+\cdots+2005k}_{\bf 2005~termeni} + \underbrace{0+1+2+\cdots +2004}_{\bf 2005~termeni}=\\\\\\ 2005 \times 2005k + \frac{2005(2004+0)}{2} =\\\\ 2005 \times 2005k + \frac{2005\times2004}{2} =\\\\ 2005 \times 2005k + 2005\times 1002 = \boxed{\bf 2005\Big(2005k + 1002\Big)~ \vdots~ 2005}[/tex]