Task B-3.5. A natural number is written on each side of the cube. Each vertex of the cube is marked with a number that is equal to the product of the numbers written on the sides of the cube that meet at that vertex. The sum of the numbers by which the vertices of the cube are thus marked is 3333. Determine the sum of the numbers by which the sides of the cube are marked.
## Solution.

Let $a, b, c, d, e$ and $f$ be the numbers written on the faces of a cube, such that $a$ and $f$ are on opposite faces, $b$ and $d$ are on opposite faces, and $c$ and $e$ are on opposite faces.

Then we have:

$$
\begin{aligned}
3333 & =a b c+a b e+a c d+a d e+b c f+b e f+c d f+d e f= \\
& =a(b c+b e+d c+d e)+f(b c+b e+d c+d e)= \\
& =(a+f)(b(c+e)+d(c+e))= \\
& =(a+f)(b+d)(c+e)
\end{aligned}
$$

Since $3333=3 \cdot 11 \cdot 101$ is the prime factorization of 3333 and each of the numbers $a+f$, $b+d$, and $c+e$ is greater than 1, it follows that $\{a+f, b+d, c+e\}=\{3,11,101\}$. Therefore,

$$
a+b+c+d+e+f=3+11+101=115
$$

## NATIONAL MATHEMATICS COMPETITION

## 4th grade - high school - B variant

## Opatija, April 1, 2011.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE THAT APPROACH AND GRADE IT ACCORDINGLY.