
2. 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 $T$.


## Solution:

Let $A B C D P Q R S$ be a cube, and the numbers $a, b, c, d, e, f$ be written on the faces $A B C D, B Q R C$, $P Q R S, A P S D, A B Q P, C R S D$ respectively. Then the products written at the corners $A, B, C, D, P$, $Q, R, S$ are respectively $a d e, a b e, a b f, a d f, c d e, b c e, b c f, c d f$. The sum of these 8 numbers is:

![](https://cdn.mathpix.com/cropped/2024_06_05_4a7377fb54a6a7232231g-1.jpg?height=585&width=528&top_left_y=1575&top_left_x=880)

This is given to be equal to $2004=2^{2} \cdot 3 \cdot 167$. Observe that none of the factors $a+c, b+d, e+f$ is equal to 1 . Thus $(a+c)(b+d)(e+f)$ is equal to $4 \cdot 3 \cdot 167,2 \cdot 6 \cdot 167,2 \cdot 3 \cdot 334$ or $2 \cdot 2 \cdot 501$. Hence the possible values of $T=a+b+c+d+e+f$ are $4+3+167=174,2+6+167=175,2+3+334=339$, or $2+2+501=505$.

Thus there are 4 possible values of $T$ and they are $174,175,339,505$.
