a+b+c=40
Fie x,y,z cele trei numere consecutive:
a/2+2 = x
b/2+6 = x+1, unde y = x+1
c/2+8 = x+2, unde z = x+2
a/2 = x-2
b/2 = x-5
c/2 = x-6
a/2+b/2+c/2 = (x-2) + (x-5) + (x-6)
(a+b+c)/2 = 3x-13
20 = 3x-13
x=11 a/2+2 = 11 a=18
y=x+1 y=12 b/2+6=12 b=12
z=x+2 c/2+8=13 c=10
a+b+c=40
18+12+10=40
QED