ExtBasic 11

[FutureHypoon]

This was most definitely my favorite mission so far. I must admit that without this thread I would not have gotten it, although (due to past experience writing scripts and progs working with large numbers) I got it as soon as mutants_r_us_guild mentioned this:

you'll notice that after a certain point the number won't be the logical product of the corresponding primes, sometimes even resulting in a negative hash.

Solved it entirely using python, although I had to install 3.1.1 in order to handle the large numbers more easily. I wish Apple would really get on the ball with their open source software and update it more often, but that's why I have Linux. We'll see what Snow Leopard brings.

Again, very fun, thanks a lot.

[chupacabras12]

This is really an interesting challenge. At one point, after you have understood how the hash is made and why you can have "strange" values for the hash, you know you must be looking for an "offset". You can blindly brute-force the script to find the value of the "offset" (very bad idea), or you can kind of figure it out by thinking binary (what is the size of an integer class in many langages). Once you have found this "offset" don't think that just adding it once to the hash value in the script will give you the answer ( you can try but you'll see that the result doesn't divide the way you want it). You now need to find the right multiple of this offset that can work. You can write a program that bruteforces with multiples of the offset, but remember: you're dealing with huge integers, so you can have an integer overflow problem that will make your solution wrong (I made a bruteforcer in C, and used the unsigned long long data type to avoid this problem)

[Ward]

Nothing like mathematica (or maple or w/e your favourite math program is) to bruteforce things involving high numbers without needing to code or think much

[dzhu]

Bash supports sufficiently large numbers as well. It's a one-liner, with the use of a certain program that lambda0 mentioned. (Well, what I have isn't that smart, so you have to sit and watch the output, but it doesn't take long.)

[Slim420]

I finished this one technically before i should have. I'm not sure if its a bug or not, but the challenge will accept any combination of the letters, I have since figured out that its an anagram with a possibility of two words (that both have the exact amount and the same letters just in a different order). So if it was meant to find the actual password, it doesnt check for that it just checks to see that all of the letters in the password are there in any order.

I dont think that was too spoilerish, but still thought i should mention it.

[faazshift]

I'm not sure if its a bug or not, but the challenge will accept any combination of the letters, I have since figured out that its an anagram with a possibility of two words (that both have the exact amount and the same letters just in a different order). So if it was meant to find the actual password, it doesnt check for that it just checks to see that all of the letters in the password are there in any order.

No bug, thats how it works.

[sirvanschaik]

wrote a brute force in C++ its seems to work right but its been running for a LONG time now hope this works

edit: alright it didnt work but mabey im doing it wrong could anyone give me possible errors

[h4ck3rz]

Finally.... done!
Well, in order to finish this mission, I think we have to know how "modulo" works. My mistake when coding the program is with the modulo stuff.

Hmm, modulo. Simple explanation about modulo is this :
If a statement a congruent b modulo m is given to us,
it means that a = nm + b, where n is a "random" value we don't know.

For example, in the mutant_r_us_guild's article,
8 wrapped by a modulo of 5 would be 3.
In this case: a = 8, m = 5, and b = 3;
Because a = nm + b --> so 8 = 5n + c;

In this case, we got the n = 1.

For another example, also in mutant_r_us_guild's article
26 wrapped by a modulo of 5 would be 1,
In this case a = 26, m = 5, and b = 1.
because a = nm + b, so
26 = 5n + 1 --> 5n = 25 --> n = 5;

We got the n here equal with 5.

Now, in this mission, we don't know what n value and a value is, but we got the b. Fortunately, b value and m value is enough to search for a and n. For the m value, like what mutant_r_us_guild said in his article, we must search for it ourself (Actually, it's "hidden" here somewhere in this thread. Just search for mutant_r_us_guild's post one by one, and you'll get the m).
But after we get that m value, it's not too hard to search for a and n. Let me give one last example :

Let's assume that b = 3; m = 7;
a = nm + b --> a = 7n + 3;
In order to get the a value, let's just brute-force the n.
If n = 1 then a = 10,
if n = 2, then a = 17,
If n = 3, then a = 24
and so on.

So, what is the relation of this modulo, the a value, n value, and another values with this mission? I won't explain about that, I'm afraid you'll be more confused . Just read the mutant_r_us_guild's article for information about that.

[raddy1313]

Finally solved it, MATLAB for the win again! Very cool mission, though I most likely would not have been able to solve it had it not been for mutant's guide. There's not really anything else I can hint that hasn't been mentioned before so just look at all the posts and read mutant's guide.

[UmniiPingvin]

Hello, I'm having a little problem with this one.
I found a combination that goes through verification in windows command line, but this mission doesn't count it as correct.
Can anyone help?
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Upd: already found correct password, sorry for bothering