Difficulty assessment: 0

Solution:

The answer is $n - 2$. 

Throw the i-th bomb at pillar 2i % 7. After each throw
you gain one position where you are sure the monster cannot be. When only 3
positions are left a single bomb kills the monster for sure. So the total
number of bombs is n - 3 + 1 = n - 2: n - 3 to get three pillars left 
and the last bomb to finish the monster.

It is not hard to see that this is optimal. Throwing bombs consecutive bombs
whose explosion is not adjacent yields not information. Neither does throwing at i % 7
instead of 2i % 7.
