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This should be fun...
Also, you should probably fix that smileyface in the first spoiler.
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On May 24 2010 04:21 TwilightStar wrote: This should be fun...
Also, you should probably fix that smileyface in the first spoiler.
Might be a part of the puzzle!
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Fuck, I'm too stupid to do this.
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You're so mean to people :O And terran too :X
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eh... i did this wrong, I dont understand part 1.
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Does anyone understand what to do with the pointer in part 1? ( dont really care about beta key, puzzle is bugging me )
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You just gave me Vietnam-style flashbacks to my year 12 4-Unit Mathematics class. Fuck. Memories of getting my dumb ass chewed out at 7am high-level maths classes returning.
Somebody help me I'm drowning
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Got it. Explanation in a little bit.
+ Show Spoiler [solution] +Part 2 is easy with an online listing of primes + Python. Part 3 is just a fancy way of saying "take the last nine digits of x1 and write them in the appropriate slots". So all we need are the a[i]. I have no clue what that smiley thing meant at all. But the listing looks like words spelled out. And there aren't many unique letters there. So its a substitution cipher. I notice right away that the word "hundred" fits in the last clue. OK. But why should I manually do this? Just plug the text into Decrypto and I get "ONE ONE TWO TEN TWENTY SEVEN TWO HUNDRED FIFTY FOUR", meaning we have a = [1,1,2,10,27,254] You see the first part of the spoiler then decrypts to "?ART ONE", obviously as a check. But that smiley man.... Now the rest is fairly simple. a = [1,1,2,10,27,254] b = [sum(a[:i+1]) for i in range(len(a))] #because we need the partial sums p = [prim[i-1] for i in b] #0-indexing versus 1-indexing x1 = reduce(lambda a,b:a*b,p,1) #multiply all elements of p together x1 is then 624888642
Nice spontaneous puzzle.
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Already claimed Congrats narri.
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On May 24 2010 05:57 narri wrote:Got it. Explanation in a little bit. + Show Spoiler [solution] +Part 2 is easy with an online listing of primes + Python. Part 3 is just a fancy way of saying "take the last nine digits of x1 and write them in the appropriate slots". So all we need are the a[i]. I have no clue what that smiley thing meant at all. But the listing looks like words spelled out. And there aren't many unique letters there. So its a substitution cipher. I notice right away that the word "hundred" fits in the last clue. OK. But why should I manually do this? Just plug the text into Decrypto and I get "ONE ONE TWO TEN TWENTY SEVEN TWO HUNDRED FIFTY FOUR", meaning we have a = [1,1,2,10,27,254] You see the first part of the spoiler then decrypts to "?ART ONE", obviously as a check. But that smiley man.... Now the rest is fairly simple. a = [1,1,2,10,27,254] b = [sum(a[:i+1]) for i in range(len(a))] #because we need the partial sums p = [prim[i-1] for i in b] #0-indexing versus 1-indexing x1 = reduce(lambda a,b:a*b,p,1) #multiply all elements of p together x1 is then 624888642 Nice spontaneous puzzle.
Excellent work! In fact, someone PMed me a correct solution almost an hour before, but luckily for you, he already had a key and was kind enough to leave it for you. Congrats!
I guess the answer to Part 1 was too easily guessed. Though I'm sure not many people are still interested in the puzzle since it's been solved, I'll give a hint for Part 1:
+ Show Spoiler [Hint to Part One] +
Thanks to everyone who participated!
EDIT: The smiley was intentional.
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