Any good password "problem solver people" out ther

[bananapeel]

I need an answer to the following question:

The red team placed one rank away from which mascot?

This is part of a puzzle game I'm involved in for the week. The answer is used to access a secret page. Please take a look at the website where this question is from: http://www3.telus.net/key2005/tues2.html

Any help or guidance is much appreciated. The other clues have already been solved by my team.

[bananapeel]

We thought red team referred to Calgary Flames, and since they lost to the Tampa Bay Lightnings they were one rank away.

The Lighning's mascot is ThunderBug, but that doesn't seem to be the right answer.

[Garrick]

red team, GO!

[ian]

Looking at the page source code, it wouldn't be possible to get it from the html, there is a javascript function that converts the answer you input to a number then compares it to the answer which is also encrypted in the same way!

Clever stuff! Sorry to be of no help though!

[MJ]

^ Ian, if i give you a fiver will you put this in your sig?

`I suck`

[bananapeel]

Looking at the page source code, it wouldn't be possible to get it from the html, there is a javascript function that converts the answer you input to a number then compares it to the answer which is also encrypted in the same way!

Clever stuff! Sorry to be of no help though!

We were thinking that it had something to do with ascii as sent to me by one of my team members

"A buddy of mine is helping me through the logic here... I've tried multiple entries but haven't got anywhere. To hack it out is difficult, but gives us a potential idea. The idea here is that you take the ascii value of the letter (lower case only, you are right) and multiplied together, the word needs to equal 2.8 e26. I've attached an ascii table link. I also attached a version of the site that returns the value of the word you just typed in instead of the word "wrong". For example, LightningBug returns 2.2 e24

http://www.asciitable.com/

Well, I can tell you that if every letter was 'm' (the middle letter
== median value) then there are probably about 13 letters in the word.
110 ^ 13 = 3.45e26.

If every letter was a 'z' 12 letters wouldn't be enough. If every
letter were an 'a' 14 letters would be too many.

This strongly leads me to believe that there are 13 letters in the
word. (If there are numbers or anything it throws major loops into it,
all bets are off. I guess it is guess and check... "

[fishingnut]

what do get to see?

[bananapeel]

the next clue to figure out in this puzzle game.

the source code, I believe, shows that the hidden page has the the following ending: password+".html"

[ian]

^ Ian, if i give you a fiver will you put this in your sig?

`I suck`

That'll be ?6 please, unless you want Isuck? To save confusion it might be worth shelling out another couple of quid to put "MJ Sucks", as being in my sig people might mistake it as refering to me! :P

[skahound]

the next clue to figure out in this puzzle game.

the source code, I believe, shows that the hidden page has the the following ending: password+".html"

But ultimately what is the prize? Is it a chance to interview with Google or something?

[picklemonkey]

there's no chance at cracking it, because it converts the (lowercased) inputted password into a 27-digit long number and checks to see if the number is correct. the onyl way to crack it is to create something to cycle through all possible combinations on the keyboard, and convert them to the same type of number, and check the two.

the number is generated by the lowercased letter ascii values. I'm guessing that the password answer has a minimum of 12 characters, since a password of all Zs would equal 10872213398722912904482816, or 1.087e25, and a password of all As would equal 67302709016557486028618977, or 6.730e26, or a maximum of 13 characters. I guess. It'd take foreeever to run a program to generate every possible combination of a 12, 13, and 14 letter word, and test it against that.

I suggest you just stick to the scavenger hunt. Then again, I could be totally wrong.