MHB Sum of Numbers on Cube Faces to Equal 2004

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Positive integers are assigned to the faces of a cube, and the product of the numbers at each vertex is calculated, with their total summing to 2004. The goal is to determine the possible values of T, the sum of the numbers on the cube's faces. The relationship between the vertex products and the face sums is crucial for solving the problem. Various mathematical approaches, including factorization and algebraic manipulation, are discussed to derive the values of T. Ultimately, the problem invites exploration of integer combinations that meet the specified conditions.
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Positive integers are written on all the faces of a cube, one on each. At each corner (vertex) of the cube, the product of the numbers on the faces that meet at the corner is written. The sum of the numbers written at all the corners is 2004. If $T$ denotes the sum of the numbers on all the faces, find all the possible values of $T$.
 
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\begin{tikzpicture}[
face/.pic = {
\node {#1};
\draw (-1,-1) rectangle (1, 1);
},
cube/.pic = {
\draw (0,0) pic {face=a};
\draw (1,-1) -- (2,-0.5) -- (2,1.5) -- (1,1) node at (1.5,0.25) {b};
\draw (2,1.5) -- (0,1.5) -- (-1,1) node at (0.5,1.25) {e};
}]
\draw (0,0) pic {face=a} (2,0) pic {face=b} (4,0) pic {face=c} (6,0) pic {face=d} (0,2) pic {face=e} (0,-2) pic {face=f};
\draw (9,0) pic {cube};
\end{tikzpicture}
Let $a,b,c,d,e,f$ be the 6 faces where $(a,c)$, $(b,d)$, and $(e,f)$ are the pairs of opposing faces as shown in the drawing.
The sum of the corners is $(ab+bc+cd+da)e+(ab+bc+cd+da)f=(a(b+d) + c(b+d))(e+f) =(a+c)(b+d)(e+f)=2004$.

The list of suitable factorizations of $2004=2^2\cdot 3\cdot 167$ is $(3\cdot 4\cdot 167,\, 2\cdot 6\cdot 167,\,2\cdot 3\cdot 334,\,2\cdot 2\cdot 501)$.
We are looking for the possibilities for $T=a+b+c+d+e+f$, which is the sum of the 3 factors.
Those possibilities for $T$ are $174,\, 175,\, 339,\, 505$.
 
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