Solve this cube if you can!

I really like this problem, so lets write a solution. To start, lets label each side. To make things simple, I will call one side x, and the side opposite to it x', another side y and its opposite y', and the last two sides z and z'. This can be done in any fashion, because if you look closely, it does not affect the outcome of the products. We are told that the vertices where the sides of the cube meet are equal to the product of the numbers on each of the sides, and that the sum of all the vertices is 2004. Thus with a little work it can be shown that:

$xyz+xyz'+xy'z'+xy'z+x'y'z'+x'y'z+x'yz'+x'yz=2004$

With a little factorization, it can be shown that:

$(x+x')(y+y')(z+z')=2004$

Now, the prime factorization of 2004 is ${ 2 }^{ 2 }\cdot 3\cdot 167$

Now that we have the prime factorization, and we know that we need three factors, i.e. (x+x'), (y+y'), and (z+z'), we can solve for all possible factorizations with three numbers. These factorizations are as follow:

$4\cdot 3\cdot 167\\ 2\cdot 6\cdot 167\\ 2\cdot 3\cdot 334\\ 2\cdot 2\cdot 501$

x+x' , y+y', or z+z' must sum to each of the factors. In the case of x+x' = 1, that indicates that the value x = 0 and x' = 1, or vice versa. Since x cannot equal zero, as it must be a positive integer, we cannot include 1 in our factorization of 2004.

Now the question asks for the sum of all the sides as T, and then the sum of all possible values of T. Since we now have all the possible factorizations of 2004 with three terms, all that is left is to check that the sum of each factorization is unique, which I can assure you is the case, and then sum all of the sums. Thus:

$(4+3+167)+(2+6+167)+(2+3+334)+(2+2+501) = \sum{T}$

$174+175+339+505 = \boxed{1193}$

Let's have a brute force solution:

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PROGRAM Cube; {174 + 175 + 339 + 505 = 1193}

USES CRT;

CONST
     rg = 1200; {Adjustable for faster search!}

VAR
    a, b, c, d, e, f, p, q: LONGINT;
    Face: ARRAY[1..16197] OF LONGINT; {Just tried to maximize.}
    i, j: WORD;

BEGIN
     CLRSCR;
     i:=0;
     FOR a:=1 TO 4 DO
      FOR b:=1 TO 4 DO
       FOR c:=b TO rg DO
        FOR d:=b TO rg DO
         FOR e:=c TO rg DO
          FOR f:=a TO 7 DO
       BEGIN
            p:=(a+f)*(b*c+c*d+d*e+e*b); {(a + f)(b + d)(c + e)}
            IF p=2004 THEN
            BEGIN
                 INC(i);
                 q:=a+b+c+d+e+f;
                 Face[i]:=q;
                 WRITELN(q, ';  ', a, ' ', b, ' ', c, ' ', d, ' ', e, ' ', f, ' ', i);
            END
       END;
       WRITELN;
       FOR j:=1 TO i DO
       IF (Face[j]<>174) AND (Face[j]<>175) AND (Face[j]<>339) THEN
          WRITELN(Face[j]);
       WRITE('Completed. i  = ', i, ' p = ', p, ' q = ', q, ' ', 
                       a, ' ', b, ' ', c, ' ', d, ' ', e, ' ', f);
       READLN
END.

Answer: $\boxed{1193}$