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FiBsTeR
United States415 Posts
As the title suggests, what follows is a beta key contest in the form of a numerical puzzle (though the numbers in the title are not relevant). I will say no more about it, but I do ask that if you are fortunate enough to solve it first, please post here or PM me saying that you have redeemed the key. GL HF! + Show Spoiler [Part One] + {STY PMR :ry" s+2 = PMR s+3 = PMR s+4 = YEP s+5 = YRM s+6 = YERMYU DRBRM s+7 = YEP JIMFTRF GOGYU GPIT + Show Spoiler [Part Two] + ![[image loading]](http://dl.dropbox.com/u/4785014/b374k3yc0n7357/2.png) + Show Spoiler [Part Three] + ![[image loading]](http://dl.dropbox.com/u/4785014/b374k3yc0n7357/3.png)   | ||
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TwilightStar
United States649 Posts
Also, you should probably fix that smileyface in the first spoiler. | ||
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cyclone25
Romania3344 Posts
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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YoonHo
Canada1043 Posts
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Muirhead
United States557 Posts
And terran too :X | ||
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AoN.DimSum
United States2983 Posts
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m4gdelen4
United States416 Posts
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AoN.DimSum
United States2983 Posts
) | ||
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snotboogie
Australia3550 Posts
Somebody help me I'm drowning | ||
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narri
United States55 Posts
+ 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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wiseman500
United States29 Posts
Congrats narri. | ||
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FiBsTeR
United States415 Posts
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] + ![[image loading]](http://dl.dropbox.com/u/4785014/b374k3yc0n7357/keyboard.png) Thanks to everyone who participated! EDIT: The smiley was intentional.  | ||
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