Enter Passcode

YIPPEE

3239

323

3239

110 days later ive finally got back to this

3239

3239

3239

oh ok

3239 is the code

It's a wonderfully simple and clever little puzzle.

It would be nice if I could reset it without clearing all my cookies so that someone else can try it on my device.

1234

Solved first in a brute-force manner using Python, then using just logic. Nice puzzle :)

No download button?

Nice!

Good game lol

Aw this was fun. I got the password right, but I'll be honest, I'm still not entirely sure why my order was correct. But then after reading over jakefriend's comments, I realized there wasn't a guaranteed order or a limit on the number of allowed guesses. So I guess I got it!

@mr_SILLY  It's not a game that requires a download--just a simple logic problem.  When you've solved it, just enter the answer in the box and it will tell you if you're correct or not.  I thought it was a fun little bite size diversion.  Pretty easy, but not completely trivial either.

it turn out... you cant download this game its just to fool you 

you really fought the pascode was down here... well no

I like chocolate

Is this game

wut i cant play :(

Where is the "download" link?

how do i play this

Does it spell a word?

So much better for you to solve it yourself but here's the simple logic with no code:

The game basically says that the answers to all 5 questions must be yes or no. There exists a certain series of yes or no for which minimum passwords are available. 

I really hope you had to click read more to see this and did not get spoiled but here goes:

To minimise the 10,000 possibilities immediately, knowing the digits immediately removes many possibilities.

If we first assume The first three digits are 1,2,3 then the only possible outcomes are 1237, 1238, and 1239

But that's 3 outcomes so let's try another way. Let's assume first digit is not 1 but second and third are still 2 and 3.

Then there can be 5 possibilites for the first digit being even and 4 being odd(as it is not 1)

Therefore, assume it to be odd to mimimise possibilities.

Let us take first digit as 3, then the first 3 digits are 323 which add up to 8, therefore if the answers to the questions are no,yes,yes,no,yes; the only possible answer is 3239

A lot of comments (from a month ago...) have this assumption that's wrecking things here is that "you can determine the passcode, and that there are no redundant questions" is recursive/contributes to itself - but, in fact, it is NOT one of the hints! The hint answers depend on this rule but, when you have the hints with answers, do NOT consider this rule for determining the code itself from the hints.

This is a fun little puzzle!

Very fun little game! It was great applying the logic and now I wanna try the original board game lol

Cool

Clever, thanks! I kinda went on a hunch at the end but that turned out to be correct.

I FOUND IT

[This is for everyone wondering what that obfuscated code @Ded Cat posted was all about :)]
**SPOILERS AHEAD**

emptyList = []

# Ybbc guebhtu nyy cbffvoyr pbqrf naq zngpu gurz gb gurve uvag nafjref

# Loop through all possible codes and match them to their hint answers

for i in range(0, 10000):
    listComprehension = [int(every) for every in str(i).zfill(4)]

    # Uvagf 1 naq 4 arrq gb or snyfr sbe erqhaqnapl

    # Hints 1 and 4 need to be false for redundancy

    magicList = [False, False, False]

    if listComprehension[0] == 1 or listComprehension[0] % 2 == 0:

        continue

    if listComprehension[1] == 2:

        magicList[0] = True
    if listComprehension[2] == 3:

        magicList[1] = True

    if listComprehension[3] > sum(listComprehension[:3]):

        magicList[2] = True

    emptyList.append([listComprehension, magicList])

# Pbhag bppheeraprf bs rnpu havdhr uvag

# Count occurrences of each unique hint

myDictionary = {}

for randomSymbol, magicList in emptyList:

    magicListTuple = tuple(magicList)

    myDictionary[magicListTuple] = myDictionary.get(magicListTuple, 0) + 1

# Cevag pbqrf jvgu havdhr uvagf

# Print codes with unique hints

for listComprehension, magicList in emptyList:

    if myDictionary[tuple(magicList)] == 1:

        print(listComprehension)

# Resulting print is [3, 2, 3, 9]

Hrm. I've narrowed it down to two possible solutions and I've used all the information and neither of my possible answers are accepted. I've double-checked my work several times and can't see my mistake.

Edit: Ah, I see, the bug in the game has already been found several times. Too bad: it'd be a fantastic puzzle if it wasn't broken.

Edit edit: Nope, I'm wrong, found the flaw in my logic, see below.

I think I'm going to sleep on this before I start working my way through all the rot13 text to see if anyone has an answer. Is there a rot13 plugin for Firefox? There must be.

If anyone wants to be generous and check my work, here's the clearest way I can find to write out my chain of logic: what am I missing?

Gur ynfg uvag pbafgenvaf gur cnffpbqr fb gung rvgure nyy qvtvgf ner mreb, be gur sbhegu qvtvg vf avar naq gur svefg guerr qvtvgf fhz gb rvtug.

Edit edit: the above is wrong. This hint says that if there is a unique passcode, it must have this property. That doesn't help you answer the question of whether a particular set of hint answers rules out all the non-unique passcodes which may or may not have that property. Gah. Tricky.

Vs gur frpbaq naq guveq uvagf ner OBGU gehr, guvf nafjref gur svsgu uvag va gur nssvezngvir naq GUNG fcrpvsvrf gur svefg qvtvg, znxvat uvagf bar naq sbhe erqhaqnag. Fb gung'f bhg.

Gur svefg naq sbhegu uvag zhfg obgu or snyfr, fvapr vs rvgure vf gehr gur bgure vf erqhaqnag. Gung nyfb nafjref gur svsgu uvag va gur nssvezngvir, fb abj jr xabj gung pbapyhfviryl.

Abj jr xabj gur nafjref gb guerr uvagf, naq gung gur erznvavat gjb pna'g obgu or gehr. Vs gurl'er obgu snyfr, gura gur fhz-gb-rvtug pbafgenvag cnerf guvatf qbja n YBG, ohg gurer ner fgvyy yvxr guvegl-gjb cbffvovyvgvrf. Fb gurer'f ab havdhr fbyhgvba gurer.

Fb rknpgyl bar bs gur frpbaq naq guveq uvagf zhfg or gehr. Naq gur svefg qvtvg zhfg or na bqq ahzore terngre guna bar.

Vs gur frpbaq uvag vf gehr, gura gur guveq qvtvg pna'g or guerr, naq gur svefg cyhf guveq qvtvgf zhfg or fvk. Gung tvirf hf n fvatyr cbffvovyvgl, svir bar gjb avar.

Vs gur guveq uvag vf gehr, gura gur frpbaq qvtvg pna'g or gjb, naq gur svefg cyhf frpbaq qvtvgf zhfg or svir. Gung nyfb tvirf hf n fvatyr cbffvovyvgl, svir mreb guerr avar.