Răspuns:
Explicație pas cu pas:
fie x+x+1+x+2+...x+2006 =(2x+2006)*2007/2= (x+2003)*2007 p suma de 2007 nr.consecutive
atunci
(x+2003)*2007=2007^2007
x+2003=2007^2006
x=2007^2006-2003
x+1=2007^2006-2002
.....
x+2006=2007^2006+3
b) e mai simplu
2007^2006= ((2007)^1003)², p.p
ramane sa il descompunem pe 2007 in suma de 5 p.p .
( !nu azis si diferite!! nu a zis si nenule!!)
2007=1600+400+4+1+1=40²+20²+1²+1²=40²+20²+1²+1²+0²
deci 2007^2007=2007^2006(40²+20²+1²+1²+0²)=
+(2007)^1003)²*(40²+20²+1²+1²+0²)=
= (2007^1003 *40)²+((2007)^1003*20)²+((2007)^1003)²+((2007)^1003)²+0²
