In the hours before Abaddon 04, a series of pages from a manual was leaked and contained information crucial to the upcoming XM-influenced events.

The first page was released at 15:00 UTC with the subsequent pages releasing in one hour increments.

The last page was released at 20:00 UTC and contained the following text:

1. Up to fourteen segments may be required when using a Z divider. (b-ez)
2. A trip to the patent office may be needed, as well as trips back and forth to base. (4965)
3. Processor and firmware are provided, although only initial instructions are needed. (8080)
4. If you spell "ohms" backwards, you get "mhos". (0.00)
5. Some resistors have no parallel. (190)
6. Capacitors are typically round, while resistors are typically rectangular cylinders.
7. Colors may mean different things, based on their position. (6.58k)
8. Consider hexadecimal values for too long may lead to floating away. (42da)
9. To display an A on a 16 element display requires 8 lit bits. (19007)
10. From an initial position of 0, A leading B, is a positive thing, but B leading A is negative.
11. A trip to base may result in characters a bit shorter than usual. (01 b0)
12. Remember that our corporate parent, Wa Bu N, is represented as .-. .. -- -- -.-.. (ヱ)
13. Please take care when removing pages from this manual. Chads can be left on or off.
14. Cryptographic pads should only be used one time, but different pads may be used sequentially. (OTPU)
15. Once all pairs are dotted, they should dash into the smallest size display (OH PS)


Thanks to Jessen Yu (@jestr)Aza Rine (@az32612), and 井伊狐狸 (@EcoliW3110) for contributing solutions to the puzzles.

All puzzles are in the volatile passcode format:

[a-z]{8}[0-9]keyword[0-9]

## 1. b-ez

b-ezabefgGkzbcefkmza-fzacdGhzbchkzbcefhkzbcefgGzajmza-dGzabefgGkza-dflzdefzabcefgGzbcefkmzacdfgG


Delimiting the string at ‘z’:

b-e
abefgGk
bcefkm
a-f
acdGh
bchk
bcefhk
bcefgG
ajm
a-dG
abefgGk
a-dil
def
abcefgG
bcefkm
acdfgG


The hint mentions fourteen segments, which would imply a 14-segment display.

The dashes mean the all the segments in between the two letters, inclusive.

bcde
abefgGk
bcefkm
abcdef
acdGh
bchk
bcefhk
bcefgG
ajm
abcdG
abefgGk
abcdil
def
abcefgG
bcefkm
acdfgG


JRWOSVNH73RDLAW5

## 2. 4965

4965628/jhqx+3zxp7ll2ixxu


The hint mentions patent office and trips back and forth to base.

4965628 refers to the US patent for Photographic Film with patent image multi-field bar code and eye-readable symbols assigned to Kodak, also known as DX encoding

Converting the remainder of the string to a binary string from Base-64:

111111 100011 100001 101010 110001 111110 110111 110011 110001 101001 111011 100101 100101 110110 100010 110001 110001 101110


The binary string needs to be inverted

000000 011100 011110 010101 001110 000001 001000 001100 001110 010110 000100 011010 011010 001001 011101 001110 001110 010001


Interpreting each of those codes as a 1st row DX contact:

000000 5000/38°
011100 40/17°
011110 32/16°
010101 100/21°
001110 64/19°
000001 3200/36°
001000 1250/32°
001100 80/20°
001110 64/19°
010110 125/22°
000100 320/26°
011010 500/28°
011010 500/28°
001001 800/30°
011101 25/15°
001110 64/19°
001110 64/19°
010001 1600/33°


Taking the degree values and converting it into binary values:

100110 010001 010000 010101 010011 100100 100000 010100 010011 010110 011010 011100 011100 011110 001111 010011 010011 100001


And finally, converting the binary string into a Base-64 string:

mRQVTkgUTWaccePTTh
=~
mrqvtkgu2accept3


## 3. 8080

8080f4d3f3fbc5844000d3d3f3eb01fbf9f9d300fbfb84f38402fb

The hint mentions processor, and 8080 is an Intel processor.

Each hex pair refers to an instruction:

F4: CP
D3: OUT
F3: DI
FB: EI
C5: PUSH
40: MOV
00: NOP
D3: OUT
D3: OUT
F3: DI
EB: XCHG
01: LXI
FB: EI
F9: SPHL
F9: SPHL
D3: OUT
00: NOP
FB: EI
FB: EI
F3: DI
02: STAX
FB: EI


Removing all except the first letter of all instructions:

CODEPAMNOODXLESSONEEADASE

CODE PAMNOOD X LESS ONE E ADA SE


X LESS ONE: W
E: 5
SE: 7

PAMNOODW5ADA7


## 4. 0.00

0.0083333333
0.0097087378
0.0090909090
0.0081967213
0.0084745762
0.0088495575
0.0097087378
0.0090909090
0.0178571428
0.0103092783
0.0100000000
0.0084745762
0.0103092783
0.0090909090
0.0101010101
0.0099009900


For each value, take the reciprocal

1 / 0.0083333333 ~ 120
1 / 0.0097087378 ~ 103
1 / 0.0090909090 ~ 110
1 / 0.0081967213 ~ 122
1 / 0.0084745762 ~ 118
1 / 0.0088495575 ~ 113
1 / 0.0097087378 ~ 103
1 / 0.0090909090 ~ 110
1 / 0.0178571428 ~ 56
1 / 0.0103092783 ~ 97
1 / 0.0100000000 = 100
1 / 0.0084745762 ~ 118
1 / 0.0103092783 ~ 97
1 / 0.0090909090 ~ 110
1 / 0.0101010101 ~ 99
1 / 0.0099009900 ~ 101


The last value is not in the tables, but a note next to the table suggests “[inserting] a 0.02S resistor to disable channel noise correction entirely.”

1 / 0.02 = 50

120 103 110 122 118 113 103 110 56 97 100 118 97 110 99 101 50


Converting this from decimal to ASCII:

xgnzvqgn8advance2


## 5. 190

190 || 245 -- 150 || 283
155 || 290 -- 190 || 230
210 || 259 -- 190 || 230
230 || 230 -- 200 || 240
100 || 117 -- 150 || 275
170 || 255 -- 190 || 285
140 || 316 -- 175 || 289
110 || 106


The hint mentions resistors and parallel.

The formula to calculate the equivalent resistance is:

$\frac{1}{R_{equivalent}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

These values are resistors in parallel and to calculate the equivalent resistance:

190 || 245 = 107.0114943 ~ 107
150 || 283 = 98.0369515  ~ 98
155 || 290 = 101.011236  ~ 101
190 || 230 = 104.047619  ~ 104
210 || 259 = 115.9701493 ~ 116
190 || 230 = 104.047619  ~ 104
230 || 230 = 115         ~ 115
200 || 240 = 109.0909091 ~ 109
100 || 117 = 53.91705069 ~ 54
150 || 275 = 97.05882353 ~ 97
170 || 255 = 102         ~ 102
190 || 285 = 114         ~ 114
140 || 316 = 97.01754386 ~ 97
175 || 289 = 108.9978448 ~ 109
110 || 106 = 53.98148148 ~ 54

107 98 101 104 116 104 115 109 54 97 102 114 97 109 54


Converting this from decimal to ASCII:

kbehthsm6afram6


## 6. Resistors/Capacitors

The first page has an image of a breadboards with resistors and capacitors on it:

• ••-• -
•-- •••
••- --•-
•- -••••
•- --• •-
•• -• ••---


Converting this from morse code gives:

eftwsuqa6again2


## 7. 6.58k

6.58kΩ ± 0.25%
7790MΩ ± 1%
6660MΩ ± 0.05%
7660MΩ ± 0.25%
85.6MΩ ± 0.1%
1690MΩ ± 0.05%
8.47kΩ ± 0.05%
2690MΩ ± 0.05%


Interpreting these as resistors and converting them into their colour equivalents:

658 16
779 71
666 78
766 76
856 96
571 69
788 48
269 68


Pairing the numbers up:

65 81 67 79 71 66 67 87 66 76 85 69 65 71 69 78 84 82 69 68


Converting the numbers from decimal to ASCII:

AQCOGBCWBLUEAGENTRED

BLUE: 6
RED: 5

AQCOGBCW6AGENT5


## 8. 42da

42da3a5e42e4364642d434bc42f03df44254635442d831aa42e435c342c6199a


The hint mentions hexadecimal, long, floats, so converting hexadecimal values to IEEE 754 floating point values:

42da3a5e 109.113998
42e43646 114.106003
42d434bc 106.102997
42f03df4 120.121002
42546354 53.097000
42d831aa 108.097000
42e435c3 114.105003
42c6199a 99.050003


Rounding each number to 3 decimal places:

109.114
114.106
106.103
120.121
53.097
108.097
114.105
99.050

109 114 114 106 106 103 120 121 53 097 108 097 114 105 99 050


Converting decimal to ASCII values:

mrrjjgxy5alaric2


## 9. 19007

S/N: 19007.21504.9420.9228.9228.763
PKI: 243.9159.959.975.240.1011.46080
PIN: 975.9420.18495.1011.9159.1019


The hint mentions 16 element displays. Each value is less than 2^16 = 65536, so converting each number into a 16-bit binary value:

0100101000111111
0101010000000000
0010010011001100
0010010000001100
0010010000001100
0000001011111011
0000000011110011
0010001111000111
0000001110111111
0000001111001111
0000000011110000
0000001111110011
1011010000000000
0000001111001111
0010010011001100
0100100000111111
0000001111110011
0010001111000111
0000001111111011


Converting the 1’s into lit segments of a sixteen segment display:

BYNVVGCR9ALEXANDER6

## 10. A/B

No known solution.

## 11. 01 b0

01 b0 c4 81 a6 c5 81 a0 5f 01 b7 5f
01 b1 05 69 b7 06 01 b0 df 19 a6 c2
14 10 85 80 06 c6 01 b0 de 69 b0 c6
69 b7 9f 81 a0 43 79 f6 9b 75 e1 9a
6d e8 1a 6c 31 1f 00 10 84 6c 27 57


The hint mentions converting to bases, converting the hexadecimal string into a Base-64 string:

AbDEgabFgaBfAbdfAbEFabcGAbDfGabCFBCFgAbGAbDeabDGabefgaBDefabdeGabegabDEfABCEbCdX


Splitting the string at intervals of A to G:

AbDEg
abFg
aBf
Abdf
AbEF
abcG
AbDfG
abCF
BCFg
AbG
AbDe
abDG
abefg
aBDef
abdeG
abeg
abDEf
ABCE
bCd


Each string segment represents a 7-bit values represented by ABCDEFG, with the values in the string being 1 if they are present, 0 if they are not.

1101101
1100011
1100010
1101010
1100110
1110001
1101011
1110010
0110011
1100001
1101100
1101001
1100111
1101110
1101101
1100101
1101110
1110100
0111000


Converting from 7-bit binary to decimal:

109 99 98 106 102 113 107 114 51 97 108 105 103 110 109 101 110 116 56


Converting decimal to ASCII:

mcbjfqkr3alignment8


## 12. ヱ

ヱミモルンヒテツルヱソヰサユホノヌルフツフウユ

The hint mentions Wa Bu N, which is a reference to Wabun Code (Japanese Morse Code)

Converting from Katakana to Wabun Code:

ヱ •--••
ミ ••-•-
モ -••-•
ル -•--•
ン •-•-•
ヒ --••-
テ •-•--
ツ •--•
ル -•--•
ヱ •--••
ソ ---•
ヰ •-••-
サ -•-•-
ユ -••--
ホ -••
ノ ••--
ヌ ••••
ル -•--•
フ --••
ツ •--•
フ --••
ウ ••-
ユ -••--


Taking the longs as 1 and shorts as 0:

01100 00101 10010 10110 01010 11001 01011 0110 10110 01100 1110 01001 10101 10011 100 0011 0000 10110 1100 0110 1100 001 10011


Combining and grouped into blocks of 8

01100001 01100101 01100101 01100101 01101101 01100110 01110010 01101011 00111000 01100001 01101100 01101100 00110011


Converting from binary to decimal

97 101 101 101 109 102 114 107 56 97 108 108 51


And finally, decimal to ASCII:

aeeemfrk8all3


## 13. Page rip

On pages 2 and 3, the pages are ripped.

^__^_^_^ ^__^__^_ ^___^__^ ^__^^__^ ^__^^__^ ^__^____ ^____^_^ ^__^____ ^^___^^_ ^__^^^^_ ^__^__^_ ^__^____ ^__^___^ ^__^^___ ^___^_^_ ^___^^__ ^^__^___


Interpreting this as a binary string:

01101010 01101101 01110110 01100110 01100110 01101111 01111010 01101111 00111001 01100001 01101101 01101111 01101110 01100111 01110101 01110011 00110111


Converting from binary to decimal:

106 109 118 102 102 111 122 111 57 97 109 111 110 103 117 115 55


Converting from decimal to ASCII:

jmvffozo9amongus7


## 14. OTPU

1. OTPUNSKCERXFTUACUXTSRG
2. CGTBIHLHNBAXLGSAVQSRQN
4. TMLQEDDVFPSNVSGQGOFSVV


The hint mentions cryptographic pads, which means using vigenere across these strings. The correct order (or transitive equivalent) to vigenere these pads is:

3 + 2 – 1 – 4

With the + being encrypt, and – being decrypt:

3 + 2:

CGTBIHLHNBAXLGSAVQSRQN
======================
OMSBLOXPBOMSBECWRPGQTU

3 + 2 - 1:
OMSBLOXPBOMSBECWRPGQTU
OTPUNSKCERXFTUACUXTSRG
======================
ATDHYWNNXXPNIKCUXSNYCO

3 + 2 - 1 - 4:
ATDHYWNNXXPNIKCUXSNYCO
TMLQEDDVFPSNVSGQGOFSVV
======================



## 15. OH PS

OH PS SL FR HD WB GH BW XD SB OC OH OX JS VS PU OH LS


Each pair of letters represents a set of 7 Morse tones.

--- ••••
•--• •••
••• •-••
••-• •-•
•••• -••
•-- -•••
--• ••••
-••• •--
-••- -••
••• -•••
--- -•-•
--- ••••
--- -••-
•--- •••
•••- •••
•--• ••-
--- ••••
•-•• •••


Interpreting these as segments of a seven-segment-display, with the segments lit for the short tones.

tEgZgFbJ4ArtiFACt6