To solve the problem of determining the minimum number of Blue beads in any configuration, we must adhere to the given rules while arranging the beads. We have a 5x5 grid resulting in 25 cells. Each bead must be Red, Blue, or Green. The constraints are:
To minimize the number of Blue beads, we must consider the maximum use of Red and Green beads under these constraints. Analyzing the scenario:
Given the structure of a row or column:
Applying the constraints:
After trying logical patterns and minimal configurations ensuring these constraints are always followed, using at least six Blue beads is essential to satisfy all constraints across a grid.
| R | G | B | G | R |
| G | B | G | R | G |
| B | G | R | G | B |
| G | R | G | B | G |
| R | G | B | G | R |
This configuration, with strategic placement of Blue beads, ensures conditions are met minimally, thereby demanding at least 6 Blue beads. Hence, the minimum number of Blue beads in any configuration 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: