6. The most common骰子is a six-sided die, it is a cube, with 1 to 6 dots on its 6 faces, and the sum of the dots on opposite faces is always 7. Now, looking at a die from a point in space, the smallest sum of all visible dots is 1, and the largest is 15 $(15=4+5+6)$. Among the numbers $1 \sim 15$, the sum of the dots that cannot be seen is $\qquad$ .

Note: The term "骰子" is translated as "die" in this context.
【Solution】Solution: The numbers on opposite faces of a die are $(1,6),(2,5),(3,4)$. From a point in space looking at a die, one might see only one face of the die, or two adjacent faces, or even three adjacent faces. Among the numbers $1 \sim 15$, the numbers $1 \sim 6$ can obviously be seen. $7=1+2+7,8=6+2,9=6+3,10=6+4,11=6+5,12=6+2+4,14=6+5+3,15=4+5+6$, and 13 cannot be decomposed, meaning that among the numbers $1 \sim 15$, the sum of the numbers that cannot be seen is 13.

Therefore, the answer is 13.