There were 3 codes found on PAC’s Weekly Wrap-up post for April 10, 2016 on the Investigation Blog.

`[1] #%$&%)$*#$$%#^%)#^%^`

[2] lojlvlyr.rzbollpy

[3] 5664277766685664339911233393335776668222

# Code #1

## Observations

The length of the code has an even number of characters, so looking at pairs of characters for a clue:

`#% $& %) $* #$ $% #^ %) #^ %^`

There are 10 groups, which could match a prefix/suffix pair with an unknown keyword. Assuming the first 5 pairs are the prefix and the last 5 are the suffix, we see that the positions where the numbers would be all have `#`

as their first character.

[hint]

Convert the symbols to their QWERTY keyboard equivalent:

`35 47 50 48 34 45 36 50 36 56`

[/hint]

Spoiler

Taking the pairs as hexadecimal numbers, we get:

`5GPH4E6P6V`

With a missing keyword. Since we converted from symbols, symbols is a candidate for the keyword:

`5GPH4SYMBOLSE6P6V`

# Code #2

Our thanks goes out to **kon5000** for giving us “pointers” to the keyword.

## Observations

Looking at the code and trying the standard methods does not reveal much, except that the `.`

seems to be in the spot of a keyword.

[hint]Since there are no numbers in the code, try transforming the letters into something else.

[/hint]

Spoiler

Remove the

`.`

and convert to

Morse
`.-.. --- .--- .-.. ...- .-.. -.-- .-. .-. --.. -... --- .-.. .-.. .--. -.--`

Make groups of 6, ignoring the spaces between the morse

`.-..-- -.---. -..... -.-..- .--.-. .-.--. .-...- --.-.. .-...- -.-.--`

Convert the code to Braille where

`-`

is a

`dot`

and

`.`

is blank and writing

`.-..--`

123456

as

.. 14

-- 25

.- 36

Result should look like this

`4 n a u 6 j 5 f 5 z`

Converting from Braille Shorthand to text, we get

`4nau6j5f5z`

Here we have a prefix and suffix, now back to the keyword.

`.`

can also be read as `Point`

, and we know one subject this with Point in its name: Darsana Point^{[1][2][3]}

`4nau6DARSANAj5f5z`

# Code #3

## Observations

The letter distribution appears random but there is a disproportionate amount of repeated numbers.

[hint]

`5 66 4 2 777 666 8 5 66 4 33 99 11 2 333 9 333 5 77 666 8 222`

[/hint]

Spoiler

First convert the groups of numbers into number pairs, starting with the number of times the number is repeated

`15 26 14 12 37 36 18 15 26 14 23 29 21 12 33 19 33 15 27 36 18 32`

On a qwerty keyboard, with rows labelled 0 to 3 for clarity we have

`0 1234567890`

1 qwertyuiop

2 asdfghjkl;

3 zxcvbnm,./

Taking the first number in each pair as the row number and the second number in each pair as the column number we get

`thrwmnithrdlawcoctjnix`

Converting th to 3, ni to 9 and oct to 8 we get

`3rwm93rdlawc8j9x`

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