To solve this problem of arranging colored beads following the given rules and maximizing the number of Red beads, we need to break down the problem systematically:
(1) According to Rule 1, no two adjacent beads can have the same color. This implies a checkerboard pattern limitation across rows and columns.
(2) Rule 2 requires at least one Green bead between two Blue beads, ensuring the alternation constraint remains.
(3) Rule 3 necessitates both at least one Blue and one Green bead between two Red beads, increasing color diversity.
Given placements: Red at 'second row, third column' and 'third row, second column'. We explore one possible optimal placement configuration maximizing Red beads within constraints:
| G | B | R | B | G |
| B | R | G | R | B |
| R | G | G | G | B |
| B | R | B | R | G |
| G | B | G | B | R |
In this configuration:
Therefore, the maximum additional Red beads possible is 6.




| A | B | C | D | Average |
|---|---|---|---|---|
| 3 | 4 | 4 | ? | 4 |
| 3 | ? | 5 | ? | 4 |
| ? | 3 | 3 | ? | 4 |
| ? | ? | ? | ? | 4.25 |
| 4 | 4 | 4 | 4.25 |
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: