## On the Subject of Beanboozled Again

When a joke goes too far...

The module shows a display with 8 encrypted characters. There are also 8 octagonal bean buttons also containing encrypted characters. In order to solve the module, press 5 beans in the right order and at the right times.

### Section 1: Decrypting the display

The display is encrypted in beanzleglyphs. Look up their corresponding base-36 character in the lookup table below.

Beanzleglyph Lookup Table | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|

1 | ||||||||||

2 | ||||||||||

3 | ||||||||||

4 | ||||||||||

5 | ||||||||||

S | 5 | Ø | J | 8 | X | |||||

O | W | Z | 1 | H | F | |||||

V | M | Q | 4 | E | L | |||||

3 | D | T | C | 6 | U | |||||

G | N | 2 | R | A | Y | |||||

P | 9 | K | I | B | 7 |

The message given is still encrypted. It is one of the words on page 2, however random letters were removed until the word length is 8, then it has been shifted randomly and then it has been encrypted using *Turntriangle Cipher* with a random coefficient array C, consisting of three coefficients ranging from 0 to 2. These values will be labeled C_{1}, C_{2} and C_{3} respectively.

#### Section 1.1: Turntriangle Cipher

Look up your character to encrypt in the cipher triangle (page 2) and call its position the pivot. Rotate your pivot clockwise 120 degrees around the center of the 6×6 triangle an amount of times equal to C1. Split the 6×6 triangle up into four 3×3 triangles. Rotate your pivot clockwise 120 degrees around the center of the 3×3 grid your pivot is in an amount of times equal to C2. Split the 6×6 triangle into nine 2×2 triangles and rotate the pivot around the center of the 2×2 triangle it is in by 120 degrees an amount of times equal to C3. Your new position is the encrypted letter.