When you get to 7 or 8 symbols, using bases doesn't work anymore. In fact, it's probably a coincidence that it did before. I've figured out a way to do them, but it's really time consuming.
Also, someone at PPT pointed out that the symbols are taken from the codestones.
The three-modifiers were only using six characters, as I recall. Though I think the "insufficient data" ones might have been using more. There were one or two error-like symbols I saw at some point.
It can't be a coincidence (well, the base whatever might be, but numerics can't): TNT had to have coded this, which means it has to follow something numerical. Which means that if it's not creating a base system, we need to find something else that replicates a base system when used with fewer numbers/symbols.
Edit: "That sounds kinda tricky... they're probably using military-grade encryption. But let's give it a shot..."
You know, I remember laughing at that originally, naively thinking it was a joke. (And sure, this isn't military or government-level encryption, but it's enough to stop the vast majority of people.)
The thing is though Rift, TNT hasn't been announcing these new puzzles in the news. I mean, you would think that if they wanted $, TNT would do that to gain more exposure. It's like this whole plot is still "underwraps".
They don't get any money from us for working on this so far: the first part had five levels, and this part was done with Flash (only the one page-view). Though it is odd...
A theory...
Facts: We have a given number and the addition don't carry.
I propose to solve row by row. I doubt we will need to solve all the 8 rows.
Then if we add 3 numbers and add the given one we have the "goal number".
Just help me to be sure...
If i need a 0 (zero), i can make it with this numbers: (0,0,0); (0,3,3); (0,2,4); (1,1,4); (0,1,5); (1,2,3); (2,2,2)
1: (0,0,1); (0,2,5); (0,3,4); (1,2,4); (1,3,3); (2,2,3)
2: (0,0,2); (0,1,1); (0,3,5); (0,4,4); (1,2,5); (1,3,4); (2,2,4); (2,3,3)
...
5: (0,0,5); (0,1,4); (0,2,3); (1,2,2); (1,1,3)
Yes, that works just fine with the two-modifiers, but with the three there are nine combinations per number (assuming my math's right). I'm very tempted to argue this isn't possible without a brute cracker, even with the knowledge we have.
On the 3 modifiers... we need an addition of 2 digits (the given and the "total").
i.e. Goal is 5; we have a 0 (zero) then we need one number... to get to five we use this combinations (0,5); (1,4); (2,3)... then we need to find the "5" missing. This means that we need to find one permutation of this ones: 5: (0,0,5); (0,1,4); (0,2,3); (1,2,2); (1,1,3).
Thats the theory...
Are you talking about just creating a "key" for how to get each symbol, or to actually solve the problems? (I should probably just get some sleep.)
I guess i need sleep too. I'm talking about solving the problem on two steps, and using just some rows (ideally just one).
1st: Goal-given
2nd: Use one combination that equals the "missing number"
EDIT: So far i'm able to find 3 adjacent rows easily... but yet not the right combination...
Here's how my friend explained it to me -- he explained it simply, and it may help some people, like me, who aren't the best at math.
C = 1, V = 2, X = 3, Z = 4, T = 5, 8 = 6.
Add up the "numbers" in a column, and then divide by 6. The remainder is what they will add up to be.
So C + V + X + Z = (1 + 2 + 3 + 4)/6 = 10/6 ... the remainder is 4. So they will add up to Z.
so in your case ... X + Z = 3 + 4 = 7. 7/6 has a remainder of 1. So it would be a C. Same thing with your minus.
I gave him my example of key: v8xzzctt and target: txcxztcz:
so, the first digit ... v + ? + ? + ? = t ... right?
2 + ? + ? + ? = 5
so now, if those 3 things add up to 3, then you get 5.
C + C + C works.
If those 3 things add up to 9 as well, that will work, since 2 + 9 = 11. And when you divide that by 6, you get a remainder of 5.
So, how can you add numbers together to get 9? 126, 135,144,234,333
So those are 5 combinations of first letters that would need to all be checked.
BUT, there is more.
if ? + ? + ? = 15, that will work as well.
since 2 + 15 = 17 ... divided by 6 is still a remainder of 5.
So 15. 366,456,555
All the numbers work this same way, though ... so in theory ... you could have a program calculate each of the combinations once you told it what the 30 lines you had to choose from were. It would just add until it found the right one.
Too much work to do by hand, if you ask me. He did one and then ran out of time before he had to sleep. Another friend also did one and it took her quite a while.
Oh, and
here is a petpage that may be helpful.
If we could pair them all will be easier... see you tomorrow. I think that someone would have a better way of solve it (i got 1 of the 3 and modifiers)