Inside the post “Susanna receives a Clue” over at the Investigate Ingress blog, there were 3 codes found inside the source and an additional one that can be seen on the image of the post.

`[1] 22122421122223132211321211131111121412132211122321141132131211111`

`[2] uyuxqbq836ojp69rzb7rq4cbqxolj4zm`

`[3] wlblwnvvlvpvdbvenfwlldvweldldvmldmwvlnmdvwvpvdmeb`

`[4] 200110001001101000011011000110010001101100011010100110011001100000011000000110000001100000011000000110000`

# Code #1

## Observations

All numbers, not inside the normal ASCII range and longer than normal

[hint]

The highest number is a 4, which would make it ideal for repeating yourself and it starts with a 2.

[/hint]

Spoiler

Perform a Run Length Encoding with the numbers, starting with 0

`0011011001111001011001100111011100110100011011010111010101101111011011100110101100111001011110100011011101101010`

This gives a nice 112 character long string, which can be nicely grouped into pieces of 8

`00110110 01111001 01100110 01110111 00110100 01101101 01110101 01101111 01101110 01101011 00111001 01111010 00110111 01101010`

Convert binary to ASCII for the final solve

`6yfw4muonk9z7j`

# Code #2

## Observations

[hint]

[/hint]

Spoiler

Convert each character to it’s Morse code

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

Converting `.`

to `0`

and `-`

to `1`

, we get something that we can once again divide nicely into 8 groups

`00110110 01100111 01100011 01111000 00111000 01110111 01101000 01111001 01100100 01100001 01101000 01101010 00110110 01111010 00111000 01110011`

Converting binary to ASCII, we get our solve

`6gcx8whydahj6z8s`

# Code #3

## Observations

[hint]

Sofar we had two binary solves, one involving RLE. Might it be that it is needed again?

[/hint]

Spoiler

Obviously we cannot run alphabet characters through a RLE cipher, but we can convert them to their alphabet index first.

Convert each character to it’s

alphabet index, where a=0 and z=25:

`22 11 1 11 22 13 21 21 11 21 15 21 3 1 21 4 13 5 22 11 11 3 21 22 4 11 3 11 3 21 12 11 3 12 22 21 11 13 12 3 21 22 21 15 21 3 12 4 1`

This looks promising, we are starting with a 2 again and no number higher than 5.

Let us group up the numbers

`221111122132121112115213121413522111132122411311321121131222211113123212221152131241`

and then perform a Run Length Encoding with them

`00110101011001000110110101101000001101110110111101110000011001010111001001100001011101000110100101110110011001010111011000110110011010000011011101100001`

What do you know, another long string that can be grouped into 8 groups

`00110101 01100100 01101101 01101000 00110111 01101111 01110000 01100101 01110010 01100001 01110100 01101001 01110110 01100101 01110110 00110110 01101000 00110111 01100001`

Finally convert binary to ASCII for the solve

`5dmh7operativev6h7a`

# Code #4

## Observations

[hint]

Look at all the information in the picture

[/hint]

Spoiler

Taking the obvious first, we see a long string of

`1`

and

`0`

, with a single

`2`

in there.

Read out the entire string, starting from the single

`2`

`200110001001101000011011000110010001101100011010100110011001100000011000000110000001100000011000000110000`

Reading the

`2`

as a hint to binary and seeing that the remaining character’s length is divisible by 8, let us group them

`00110001 00110100 00110110 00110010 00110110 00110101 00110011 00110000 00110000 00110000 00110000 00110000 00110000`

Converting Binary to ASCII we get

`1462653000000`

Recognize that it look very much like a

Unix timestamp, we use an

Timestamp Epoch convert and get the UTC date and time of

`2016-05-07T20:30:00+00:00`

So now we have a date and time, but that doesn’t mean much, lets have a look at the other clues.

The filename of image was `lccn-n96059244.jpg`

Googling a bit, we see that `lccn`

stands for Library of Congress Control Number.

Taking that knowledge, we search their database and find that `n96059244`

points to Tim Hawkinson, which corresponds with what looks to be in the image.

Over at the University of California, San Diego, there is a Stuart collection which includes the Tim Hawkinson Bear.

So it seems that something might happen at or around the bear on the 7th May 2016 at 20:30, which was confirmed later in the post “A Mission in San Diego”

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