5. Matija first wrote down all the numbers from 1 to 10000 in order, and then erased all the numbers that were not divisible by either 5 or 11. What number is in the 2004th position?

## Problems for 3rd Grade
II/5. Since the least common multiple of the numbers 5 and 11 is 55, there are exactly $11+5-1=15$ numbers divisible by 5 or 11 among the numbers from 1 to 55 inclusive. These are: $a_{1}=5, a_{2}=10$, $a_{3}=11, a_{4}=15, a_{5}=20, a_{6}=22, a_{7}=25, a_{8}=30, a_{9}=33, a_{10}=35, a_{11}=40$, $a_{12}=44, a_{13}=45, a_{14}=50$ and $a_{15}=55$. Since $2004=133 \cdot 15+9$, the desired number is $133 \cdot 55+a_{9}=133 \cdot 55+33=7348$.

Observation that it is good to observe blocks of 55 numbers: 2 points. Observation that there are exactly 15 numbers divisible by 5 or 11 between 1 and 55 (can also be implicit by listing all these numbers): $\mathbf{1}$ point. Equation $2004=133 \cdot 15+9$ or an equivalent one that indicates we need to observe the 9th number in the block: 2 points. Conclusion $133 \cdot 55+a_{9}=7348$: $\mathbf{2}$ points.