B3. Jure put 2012 black frnikols in a row. First, he replaced every third frnikol in the row with a red frnikol. Then he replaced every fifth frnikol in the row with an orange frnikol. Finally, he replaced every seventh black frnikol in the row with a blue frnikol. How many black frnikols did he have left in the row at the end?

## Problems for 3rd Grade

Time for solving: 120 minutes. In section A, a correct answer will be worth two points, while a wrong answer will result in one point being deducted. Enter your answers for section A in the left table.
![](https://cdn.mathpix.com/cropped/2024_06_07_66bdf7a61d9bfcb32ca4g-06.jpg?height=164&width=802&top_left_y=546&top_left_x=640)
B3. Let's mark the frnikoles by type with numbers from 1 to 2012. First, let's calculate how many black frnikoles were in the row after the second step. Jure exchanged exactly those black frnikoles whose number is divisible by 3 or 5. Since $2012=670 \cdot 3+2=402 \cdot 5+2=134 \cdot 15+2$,
there are exactly $670+402-134=938$ such frnikoles. Therefore, after the second step, there were still
1074 black frnikoles in the row. Since $1074=153 \cdot 7+3$, in the third step, Jure exchanged 153 black
frnikoles. In the end, he had $1074-153=921$ black frnikoles left in the row.
The calculated number of frnikoles after the first step (2012-670) ..... 1 point
The calculated number of frnikoles after the second step (2012-670-402+134) ..... 2 points
The calculated number of frnikoles he exchanged in the third step (153) ..... 2 points
The calculated final number of black frnikoles ..... 1 point

## Solutions for 3rd grade

Within section A, a correct answer will be awarded two points, while an incorrect answer will result in the deduction of one point. To avoid a potential negative final score, each contestant is given an initial 3 points.

| A1 | A2 | A3 |
| :---: | :---: | :---: |
| $\mathrm{D}$ | $\mathrm{B}$ | $\mathrm{A}$ |

Justifications: