Solution:
First let we say what A, B and C cannot be - they
cannot be all odd (beause the sum is even) and they cannot be two even + one odd (again for the same reason).
The variants left are: all even or two odd and one even
1. If all of them are even - it is obvious that their multiplication can be divided on 4:
A*B*C*D*E/4 = (A/2)*(B/2)*C*D*E => because both A and B are even - they can be divided on 2 without remainder => 1. is Prooved
2. Let we check the variant if A and B are odd and C is even. Let's see what is happening with D and E:
A+B+C+D+E = 2007 => D+E = 2007-(A+B+C) => D+E = 1007, which means that one of (D,E) is odd and one is even. => from the five numbers we have one even from (A,B,C) and one even from (D,E) => we have total 3 odd and
2 even - again like one we have (A*B*C/2)*(D*E/2) =>... prooved!
1. + 2. = PROOVED
