1+2+3+...+2004=(2004x2005)/2=1002x2005
2+4+6+...........+2004 =2(1+2+3+...+1002)=2(1002x1003)/2=1002x1003
S-a folosit 1+2+3+...+n=n(n+1)/2
(1+2+3+........+2004)/(2+4+6+...........+2004) = (2x-1)/x
(1002x2005)/(1002x1003)=(2x-1)/x
2005/1003=(2x-1)/x
2005x=1003(2x-1)
2005x=2006x-1003
1003=2006x-2005x
x=1003