Lonelygirl15

To my "Brother" :

Step 5: A tiny project.

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[ a tiny hint: jx62t ]

SurrealisticPill wrote: so i think this means we have to cut out a wheel, write the code around the edges, and cut out another smaller wheel, write the other code around the edges of that one, and attach them in the center with a brad.

should give up a decoder wheel/ring.

deagol wrote: Here's my take on this tiny project:

x=o
xx=oo (confirmation of x=o)

So, take your decoder ring and match x in one ring with o in the other ring by shifting them. Now, there's two ways of doing this:

Code: Select all

 Movies  C64FB2835D7A91E
  Ring   xtvouzwpmrnyqsl
---------|--------------
Loveline onzsymtvpqlwurx
  Ring   D61A5EF4793B8C2
         ^
         |
       match

or

Code: Select all

 Movies  C64FB2835D7A91E
  Ring   xtvouzwpmrnyqsl
------------|-----------
Loveline wurxonzsymtvpql
  Ring   B8C2D61A5EF4793
            ^
            |
          match

So, now we can decode any letter in one ring to the corresponding letter in the other.

Tiltingwindward wrote:We know that this step is called "A Tiny Project." The tiny is a reference to tinyurl, of course, which is how we found this pictures in the first place. But the project part would make a lot more sense if this was the step where we physically put together the two halves of our decoder ring.

In the chat, we've been talking about the very small ruler that is concealed near the top of the scissors picture. It's certainly not there in the original, and I think we have to assume that it's meant for us to use. But if we're building a decoder ring, then the ruler might come in handy so that we know which set of letters/numbers goes on the outside of the ring, and which set goes on the inner. That is, if x.jpeg stands for one set of letters/numbers, and the ruler shows that the diameter of the brad is smaller than the diameter of the scissors in xx.jpeg (which would stand for the other set of letters/numbers), then we would know that the x.jpeg set of letters/numbers goes on the inside of the decoder ring.

See where I'm going with this?