Algebra:
numerele consecutive au diferenta 1
a+b+c+d+e=2000
b=a+1
c=b+1=a+2
d=c+1=b+2=a+3
e=d+1=c+2=b+3=a+4
a+a+1+a+2+a+3+a+4=2000
5a+10=2000
5a=2000-10
5a=1990
a=1990:5
a=398
b=398+1=399
c=399+1=400
d=400+1=401
e=401+1=402
Metoda grafica:
numerele consecutive au diferenta 1
a+b+c+d=2000
b=a+1
c=b+1=a+2
d=c+1=b+2=a+3
e=d+1=c+2=b+3=a+4
a I......I
b I.........I+1
c I.........I+1+1
d I..........I+1+1+1
e I.....I+1+1+1+1
1+1+1+1+1+1+1+1+1+1+1=10(suma partilor cunoscute)
2000-10=1990(cinci parti egale)
1990:5=398(a)
398+1=399(b)
398+1+1=400(c)
398+1+1+1=401(d)
398+1+1+1=402(e)
