The rules say and the visuals show that the puzzle operates under Symbol Cycle logic, but with more displays and cycles with lengths higher than 5.
Notice how the penultimate number shares an element with a previous number in every set, which can be used to work out the 'modulus' for each column. These 'moduli' are as follows:
| 37 | 11 | 17 | 23 | 19 | 31 | 13 | 29 |
Moduloing each given number by these values yields:
| %37 | %11 | %17 | %23 | %19 | %31 | %13 | %29 | |
| 9367 | 6 R | 6 E | 0 S | 6 N | 0 R | 5 L | 7 R | 0 N |
| 17579 | 4 D | 1 R | 1 L | 7 O | 4 O | 2 R | 3 A | 5 T |
| 25234 | 0 M | 0 I | 6 R | 3 R | 2 L | 0 N | 1 E | 4 R |
| 69602 | 5 S | 5 U | 4 E | 4 E | 5 A | 7 E | 0 S | 2 S |
| 75925 | 1 P | 3 O | 3 N | 2 L | 1 E | 6 O | 5 N | 3 G |
| 2980685 | 2 T | 4 L | 7 O | 0 T | 3 N | 4 T | 6 O | 7 Y |
| 213267045 | 7 B | 2 N | 5 I | 5 A | 6 T | 3 I | 4 L | 1 D |
| 35120402852 | 3 C | 7 A | 2 A | 1 I | 7 I | 1 A | 2 T | 6 E |
| 35121391230 | 0 M | 2 N | 0 S | 0 T | 5 A | 6 O | 3 A | 6 E |
| 35121391231 | 1 ? | 3 ? | 1 ? | 1 ? | 6 ? | 7 ? | 4 ? | 7 ? |
Replacing the digits in the bottom row with their letter counterparts yields the solution.
POLITELY