A downloadable game

I've forgotten my 4 digit passcode!

Fortunately, I saved some hints on my phone. All the hints have a yes/no answer. I remember that if you know all the answers then you can determine the passcode, and that there are no redundant questions.

The hints are:

  • Is the first digit 1?
  • Is the second digit 2?
  • Is the third digit 3?
  • Is the first digit even?
  • Is the fourth digit larger than the sum of the first three digits?

Unfortunately, to access the answers to the hints, I need to log in to my phone with my passcode :(

Check solution here

Inspired by the boardgame Turing Machine. Thanks to Ricky Liu, Jamie Perconti, personman, and Mokesmoe for playtesting.

Rated 4.2 out of 5 stars
(22 total ratings)
AuthorDraknek & Friends
TagsAbstract, Detective, Math


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


No download button?


Good game lol


wheres the download button

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

Hint to help you get started: (hopefully you have to hit 'read more' to see this!)

Consider the premise that you should not need to bruteforce the code, at all: The questions, when answered correctly, will only ever lead you to a single possibility of numbers that can be the code.

Then, look at the fifth question first. What would have to be true for this question to indicate a singular numeric answer, between 1-9?

More direct hint in next comment.

More direct follow-up hint:

In order for you to get the fourth digit, the fifth question must indicate exactly one number, and only one number. Because it is comparing a sum to itself, the only way it can isolate a single number is at either end of the 1-9 set, if the total of the other three numbers is either less than 2 or greater than 8.

Even more direct hint in next comment.

Even-more direct follow-up hint (with one number worked out):

If the first three numbers are all 1 (as low as they can be) and the fifth question is false, then their sum is 3 and the fourth number is either 1 or 2. Because that doesn't isolate a single number, that can't be right - so the fourth number must be 9, with the fifth question being true and the sum of the first three numbers being 8.

From here, the rest of the solving should be a lot easier - there's only so many combinations of the numbers on this phone that can total 8!

Thanks for this! I got the right answer but was a little confused as to why my order was correct. Your comment that "there's only so many combinations of the numbers on this phone that can total 8" set my mind at ease that there isn't a clue necessarily for the correct order, but that a few different guessing attempts are allowed.

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

(1 edit) (+1)

Can someone please explain this, do not understand the question. People in the comments are writing such strange texts

(2 edits)

There is just one way to answer yes or no to all of those questions together, which yields one, and only one, password. 

You have to use logic to try and find a way of answering the questions so that the number of passwords that fit all criteria are the minimum possible, you find the right answer when the number of possible passwords is 1 for a given yes or no sequence.

I recommend using a notepad or something to organize the rules and try to make connections between them.

when you think you have the answer, click on "Check solution here" and input what you think is the password.

Thanks, I managed to figure it out after reading your reply. I tried using excel and python to code it all but then found the right answer with just simple logic mentally. Wonderful puzzle!

Vi7 7 All of the weird text you have been seeing is a caesar-cipher.

Oh so every letter was shifted by a certain number of characters? That's cool to know, thank you! Wjy are they doing that though? For privacy?

I Don't Know. But, probably for privacy.

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


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



(2 edits)

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

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:


    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:


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

(2 edits)

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.


Unique lil game. It definitely has a unique solution too.

Probably not the way I was supposed to solve it but here is the python script I made to find the answer :) 

(in rot13 cuz thats the trend i guess):

fbyhgvbaf = []
# Ybbc guebhtu nyy cbffvoyr pbqrf naq zngpu gurz gb gurve uvag nafjref
sbe v va enatr(0, 10000):
    pbqr = [vag(qvtvg) sbe qvtvg va fge(v).msvyy(4)]
    # Uvagf 1 naq 4 arrq gb or snyfr sbe erqhaqnapl
    uvag = [Snyfr, Snyfr, Snyfr]
    vs pbqr[0] == 1 be pbqr[0] % 2 == 0:
    vs pbqr[1] == 2:
        uvag[0] = Gehr
    vs pbqr[2] == 3:
        uvag[1] = Gehr
    vs pbqr[3] > fhz(pbqr[:3]):
        uvag[2] = Gehr
    fbyhgvbaf.nccraq([pbqr, uvag])
# Pbhag bppheeraprf bs rnpu havdhr uvag
uvag_pbhagf = {}
sbe _, uvag va fbyhgvbaf:
    uvag_ghcyr = ghcyr(uvag)
    uvag_pbhagf[uvag_ghcyr] = uvag_pbhagf.trg(uvag_ghcyr, 0) + 1
# Cevag pbqrf jvgu havdhr uvagf
sbe pbqr, uvag va fbyhgvbaf:
    vs uvag_pbhagf[ghcyr(uvag)] == 1:

How does your code work? How does it know it has guessed the right password? And is this a bruteforce crack?


Fun!  Clever use of the itch.io password system :)


I love this puzzle, but I think it's not quite correct: three solutions are compatible with the specifications. Spoilers and discussion in rot13:

Gur svefg naq sbhegu dhrfgvbaf zhfg or nafjrerq "ab", gb nibvq erqhaqnapl, naq gur svsgu zhfg or nafjrerq "lrf" gb trg rabhtu vasbezngvba nobhg gur sbhegu qvtvg, fb jr pna pbafvqre whfg uvagf 2 naq 3.

Gur npprcgrq fbyhgvba pbzrf sebz nafjrevat obgu "lrf". Nafjrevat obgu "ab" qbrfa'g tvir rabhtu vasbezngvba - vg yrnirf, rt, 3419 naq 3149 nf cbffvovyvgvrf. Ohg jr pna nafjre 2 jvgu "lrf" naq 3 jvgu "ab" naq trg n fbyhgvba bs 5219, be nafjre 2 jvgu "ab" naq 3 jvgu "lrf" gb trg n fbyhgvba bs 5039.

Obgu pnfrf yrnir ab erqhaqnag dhrfgvbaf, naq va obgu pnfrf xabjvat gur svir nafjref vf rabhtu gb svaq bar fbyhgvba.

Vs lbh xabj nyy gur nafjref gura lbh pna qrgrezvar gur cnffpbqr. Vs rvgure bs uvagf 2 be 3 ner "ab", gur nafjref gb gur uvagf jbhyq abg havdhryl pbafgenva gur ynfg qvtvg, naq gurer jbhyq or zhygvcyr cbffvoyr fbyhgvbaf.

Vs uvagf 2, 3 naq 5 ner "lrf", lbh pna nyernql qrqhpr nyy sbhe qvtvgf! Lbh'ir tbg k23l, lbh xabj l > k+2+3, naq *gur chmmyr gryyf lbh gurer vf bayl bar fbyhgvba*. Fb k+2+3 pnaabg or yrff guna 8, be gurer jvyy or zhygvcyr cbffvovyvgvrf sbe l, naq vg pnaabg or zber guna 8, be gurer jvyy or ab cbffvovyvgvrf sbe l.


> Vs rvgure bs uvagf 2 be 3 ner "ab", gur nafjref gb gur uvagf jbhyq abg havdhryl pbafgenva gur ynfg qvtvg, naq gurer jbhyq or zhygvcyr cbffvoyr fbyhgvbaf.

Naq, ntnva, gur chmmyr gryyf lbh gung jba'g unccra! Fb lbh pna nffhzr gung gur bayl jnl gur nafjref qb pbafgenva nyy gur qvtvgf vf vs gur fhz bs gur svefg guerr qvtvgf vf 8, naq gura obgu lrf/ab naq ab/lrf sbe uvagf 2 naq 3 tvir lbh rknpgyl bar fbyhgvba.

Gur rkcrpgrq fbyhgvba gnxrf nqinagntr bs guvf fnzr ybtvp: nafjref "ab, lrf, lrf, ab" gb gur svefg sbhe dhrfgvbaf yrnir cbffvovyvgvrf bs 3, 5 naq 7 sbe gur svefg qvtvg. Gur bayl ernfba lbh pna cvpx bar vf gung gur inyhr unf gb pbafgenva gur ynfg qvtvg. Ohg gur fnzr vf gehr vs lbh'ir bayl nafjrerq "lrf, lrf" gb uvagf 2 naq 3: gur svefg qvtvg pbhyq or nalguvat, ohg bayl bar inyhr jbexf.

Lbh pna guvax bs guvf nf n evqqyr jurer 2 cresrpg ybtvpvnaf jvgu qvssrerag vasbezngvbaf gnyx gb rnpu bgure ol bayl fnlvat fghss yvxr "v qba'g xabj gur nafjre" naq "ng svefg v qvqa'g xabj gur nafjre, ohg abj gung lbh gbyq zr lbh qvqa'g xabj gur nafjre, v xabj gur nafjre"
-Qenxarx pubbfrf n cnffpbqr.
-Qenxarx tvirf gur 5 uvagf naq gurve nafjref gb Nyvpr.
-Qenxarx tvirf gur 5 uvagf jvgubhg gurve nafjref gb Obo.
-Nyvpr gryyf Obo gung fur xabjf gur cnffpbqr, naq gung vs bar bs gur 5 uvagf jrer erzbirq fur jbhyqa'g xabj vg
-Lbh'er Obo, lbh unir gb svaq gur cnffpbqr
Abj ubj gb fbyir vg:
-Jvgu lbhe xabjyrqtr gung ab uvagf ner erqhaqnag, Obo pna qrqhpr gung obgu uvagf 1 naq 4 zhfg or nafjrerq ol n Ab; Qvtvg 1 = 3,5,7 be 9
-Jvgu lbhe xabjyrqtr gung Nyvpr xabjf gur cnffpbqr, Obo pna qrqhpr gung uvag 5 zhfg or nafjrerq ol n Lrf, bgurejvfr gurer jbhyq or jnl gbb zhpu cbffvovyvgvrf
-Abj vf gur cneg v guvax lbh zvffrq, vs sbe rknzcyr Uvag 2 jnf nafjrerq ol n Ab, gura Nyvpr jbhyqa'g xabj gur cnffpbqr sebz gur trg tb (sbe nyy fur xabjf vg pbhyq or nal pbqr orgjrra 3037, 3038, 3039, 3138, 3139 be 5039). Gur snpg gung fur xabjf gur cnffpbqr sebz gur uvagf nybar yrgf Obo qrqhpr gung Uvag 2 vf nafjrerq ol n Lrf.
-Gur fnzr ybtvp pna or qbar sbe Uvag 3. Abj Obo xabjf gur nafjref gb nyy 5 Uvagf (ALLAL), naq pna rnfvyl svaq gung 3239 vf gur bayl cbffvoyr cnffpbqr

Jung V guvax lbh'er zvffvat sebz zl nethzrag vf gung fvapr uvag 5 vf gur bayl cvrpr bs vasbezngvba jr unir nobhg gur sbhegu qvtvg, jr pna xabj, jvgubhg pbafvqrevat gur bgure dhrfgvbaf ng nyy, gung gur sbhegu qvtvg zhfg or 9 naq gur fhz bs gur bgure guerr zhfg or 8. Bgurejvfr gur chmmyr vf hafbyirnoyr. Fvapr jr ner cebzvfrq gung gur chmmyr vf fbyirnoyr, jr pna znxr guvf qrqhpgvba, naq nafjrevat "ab" gb uvag 2 be "ab" gb uvag 3 rnpu tvir hf rknpgyl bar fbyhgvba (naq nafjrevat "lrf" gb obgu znxrf dhrfgvbaf 1 naq 4 erqhaqnag).

Lbhe Nyvpr naq Obo fvghngvba qbrfa'g zveebe bhef rknpgyl: Nyvpr qbrfa'g xabj gur ceboyrz vf fbyirnoyr! Naq Obo xabjf fur qbrfa'g xabj guvf, naq gung gurersber fur fbyirq vg jvgubhg gur qrqhpgvba nobir. Va gung pnfr, gurer vf vaqrrq bayl bar fbyhgvba, gur rkcrpgrq bar. Ohg jr qb xabj gur chmmyr vf fbyirnoyr, naq gung punatrf gur vasbezngvba ninvynoyr gb hf naq gur cbffvoyr fbyhgvbaf. Hasbeghangryl, fvapr gjb fbyhgvbaf erznva, gur chmmyr raqf hc frys-pbagenqvpgbel.

"V erzrzore gung vs lbh xabj nyy gur nafjref gura lbh pna qrgrezvar gur cnffpbqr" vf jung znxrf gur nyvpr/obo fvghngvba zveebe guvf chmmyr rknpgyl
Nyvpr qbrfa'g xabj nalguvat nobhg gur chmmyr jr'er gelvat gb fbyir, ohg jr'er gbyq gung fur pna svther bhg gur cnffpbqr ol bayl hfvat gur 5 uvagf naq nafjref, juvpu jbhyqa'g jbex jvgu uvagf 2 be 3 orvat nafjrerq ol n ab


njvrzm, I've been enjoying the discussion here, and I've just written a little solver for this puzzle. Now I'm curious if you consider this updated puzzle broken as well:

I've forgotten my 4 digit passcode!

Fortunately, I saved some hints on my phone. All the hints have a yes/no answer. I remember that if you know all the answers then you can determine the passcode.

The hints are:

  • Is the first digit smaller than the second digit? 
  • Is the second digit smaller than the third digit?
  • Is the third digit smaller than 4?
  • Is the first digit even? 
  • Is the fourth digit smaller than the sum of the first two digits?

Unfortunately, to access the answers to the hints, I need to log in to my phone with my passcode :(

(Notes: There's no information about redundancy of the hints. Also, this puzzle has a unique solution)

no, it could be 3209, 3219, 3229, 5219, or 5209 without the 3rd questions being a yes, and 3039, 3139, or 5039 without the 2nd question being a yes. They have multiple possible codes, so you know those answers arent correct

how do i get to the input password screen after solving? After I solved it i wanna go back to that screen but the 'check solution here' link leads straight to the congrats screen


You can open the link in incognito mode, so it does not access your cookie.

i cant play??? there is no game how did you guys play it


ooo you need to click on 'check solution'

idk how to figure this out, if hint one is yes then hint 4 is no, same way other way around, the only problem is that i have no way to know wich of the two is correct, the last hint is also useless, the fourth digit could be anything. maybe im just stupid lol

i guess i just have to try it 6561 times and see wich one fits

Focus on the fact that if you know all of the answers, you can figure out the password. this means that the 2nd digit and 3rd digit has to be 2 and 3, as if it wasn't the solution wouldn't be restricted to one solution. after that, focus on the fifth hint. with logical reasoning, you should be able to figure it out. remember, you have to narrow it down to ONE solution.

This was fun! Would be curious to see what the "proper" steps were... my solve feels kinda big and clunky.

how am i supposed to find out the passcode if i d ont have the answers to the hints...

(1 edit) (+1)

This is the key part of the instructions:

I remember that if you know all the answers then you can determine the passcode, and that there are no redundant questions.

This means that if you knew the yes/no answers to each of the 5 hints, then only a single 4-digit passcode would satisfy all five of the hint answers. Additionally, it means that if you removed any one of the hints, then there would be multiple 4-digit passcodes that would satisfy all four of the remaining hint answers.

To get you started, think about the combination of hints 1 and 4. If the answer to hint 1 is "Yes", is hint 4 redundant? If the answer to hint 4 is "Yes", is hint 1 redundant?

nvm i give up i used all my math brain power (this is literally probably so easy and im so very stupid) 


I figured


how do i play 🏳‍🌈

(1 edit)

Don't know how to write in this cryptic style like some others do, but I am pretty sure I found a list of answers and can determine an unambiguous passcode from them, but this one isn't accepted. I can see the numbers in the cryptic texts, that would be included in my code.

(3 edits)

Hmmm, I can think of one thing you might be missing.

Are you also following the rule that no hint is redundant? If you follow that rule, there are two questions that have only one possible answer.

Not sure what this hint is implying. Redundant in relation to what?

No answer should should render a question unnecessary. Think about what information each question gives you. Some questions would render others irrelevant if you answered them in a particular way
(1 edit)

Very cool idea and a creative take on the detective genre!

One small niggle:

V guvax gur "ab erqhaqnag dhrfgvbaf" pbafgenvag unf n fyvtug ybtvpny synj.

Gur nafjre gb dhrfgvba 5 zhfg nyjnlf or "lrf" fvapr vg'f gur bayl dhrfgvba gung tvirf vasbezngvba nobhg gur sbhegu qvtvg naq gurer vf ab qvtvg gung pbafgenvaf gur cnffjbeq gb bar fbyhgvba vs gur nafjre vf "ab". Nyfb, fvapr guvf vf gur bayl vasbezngvba jr unir ba gur sbhegu qvtvg, jr xabj gung gur "lrf" nafjre zhfg pbafgenva gur sbhegu qvtvg gb bar cbffvovyvgl. Gurersber gur sbhegu qvtvg zhfg or 9.

Vs gur sbhegu qvtvg vf 9 naq gur nafjref gb dhrfgvbaf 2 naq 3 ner "lrf", gura dhrfgvbaf 1 naq 4 ner erqhaqnag, fvapr jr nhgbzngvpnyyl xabj gur svefg qvtvg (gur fhz bs gur svefg guerr qvtvgf zhfg or 8 gb sbepr gur ynfg qvtvg gb or 9).

Dhrfgvbaf 1 naq 4 nera'g erqhaqnag: gel fbyivat gur chmmyr jvgubhg gurz!

(1 edit)

Zl cerivbhf pbzzrag jnf zrnag gb or n fbyhgvba juvpu qbrfa'g hfr dhrfgvbaf 1 naq 4.

V thrff V unir gur fnzr pevgvpvfz nf lbh. Ubj qb jr xabj gung dhrfgvbaf 2 naq 3 ner "lrf"? Naq vs gurl ner, nf gur fbyhgvba nffhzrf, gura gung vaureragyl znxrf dhrfgvbaf 1 naq 4 erqhaqnag, guhf oernxvat gur vagraqrq ehyr.

V pnzr onpx gb rqvg zl zrffntr jvgu fbzrguvat gb gur fnzr rssrpg, zl ncbybtvrf :) Vg bppheerq gb zr gb gnxr zl bja fhttrfgvba frevbhfyl. Vs 2, 3 naq 5 jrer gur bayl uvagf, gura gur "lbh unir rabhtu vasbezngvba" cebivfb jbhyq sbepr gurz nyy gb or "lrf" naq yrnq gb n fvatyr fbyhgvba. Va gung frafr dhrfgvbaf 1 naq 4 ner vaqrrq erqhaqnag. Juvpu znxrf zr ernyvmr: lbhe cbvag zrnaf gung gur npprcgrq nafjre vf gur bayl bar bs gur guerr gung'f _abg_ pbeerpg!

Nice puzzle! Surprised we can solve it even if we don't know hint's answers.

Very cool little puzzle, thanks Draknek!

Here's my absolutely rigorous, fiercely logical and not at all intuive walkthrough for those that need a hint (rot13):

Jr arrq gb svaq n frg bs nafjref gb gur svir dhrfgvbaf fb gung gurer rkvfgf rknpgyl bar 4-qvtvg qrpvzny ahzore gung fngvfsvrf gur nafjre frg, juvyr gurer zhfg or ab erqhaqnag dhrfgvbaf.

Jr pna vzzrqvngryl frr gung gur nafjre gb dhrfgvba 1 vf Ab orpnhfr bgurejvfr dhrfgvba 4 jbhyq or erqhaqnag.

Sebz urer, yrg'f vaghvgviryl znxr rqhpngrq nffhzcgvbaf naq gel gb nffvta nafjref fb gung rnpu nafjre unf znkvzhz "vzcnpg" naq guhf gvtugraf gur fbyhgvba fcnpr sbe gur qvtvgf gur zbfg:

Yrg'f svk gur frpbaq naq guveq qvtvg naq nffhzr gur nafjref gb dhrfgvbaf 2 naq 3 ner obgu "lrf".

Jr gura xabj gung gur fhz bs gur 2aq naq 3eq qvtvg vf 5. Sebz urer, jr nffhzr gung nafjre 5 vf nyfb "lrf", fb jr gura xabj gung qvtvg 4 arrqf gb or "ovt".

Fb gur dhrfgvba vf: pna jr svaq n qvtvg 1 gung jr nqq gb 5 naq fgvyy svaq n qvtvg 4 gung'f ovttre?

Gurer ner frireny, ohg vs jr nyfb nffhzr nafjre 4 vf "ab", gura qvtvg 1 zhfg ng yrnfg or 3 (vg pnaabg or rira, naq vg pnaabg or 1). Gura jr trg 3+2+3 < qvtvg 4, fb qvtvg 4 pna bayl or 9.

Fb abj jr'ir svkrq na nafjre frg (ab,lrf,lrf,ab,lrf) gung yrnirf hf jvgu  rknpgyl bar cva ahzore: guerr gjb guerr avar

Gurer'f n ceboyrz jvgu guvf ernfbavat, naq gb zr vg oernxf gur chmmyr: gurer'f ab ernfba lbh unir gb nafjre dhrfgvbaf 2 naq 3 jvgu "lrf". Va snpg vs lbh nafjre gurz "ab", "lrf" naq "lrf", "ab" erfcrpgviryl (jvgu gur bgure nafjref erznvavat gur fnzr) lbh trg gjb qvssrerag nafjref: 5039 naq 5219. Obgu fngvfsl gur pbaqvgvbaf bs gur chmmyr: xabjvat gur nafjref gb gur dhrfgvbaf tvirf lbh gur fbyhgvba, naq erzbivat nal dhrfgvba yrnirf lbh jvgu zhygvcyr fbyhgvbaf.

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>Both satisfy the conditions of the puzzle

The winning condition is to have an answer set that implies a unique solution. Both of your suggested answer sets allow for multiple pin numbers:

Sbe  nafjre frg (0,0,1,0,1) lbhe fhttrfgrq fbyhgvba 5039 vf abg gur bayl bar: r.t. 3139 naq 3138 jbhyq nyfb or inyvq.

Fnzr sbe (0,1,0,0,1): 5219 vf abg gur bayl fbyhgvba, r.t. 3217, 3218 naq 3219 jbhyq nyfb jbex.

By the same logic, the same is true of the original puzzle!

Gur rkcrpgrq nafjref (0,1,1,0,1) qb abg ryvzvangr 3139 naq 3138 nal zber guna (0,0,1,0,1) be (0,1,0,0,1) qb. 3 vf abg 1 naq 3 vf abg bqq, fb dhrfgvbaf 1 naq 4 qb abg ryzvangr vg. Jung qrgrezvarf gur havdhr fbyhgvba vf gur fnzr va nyy pnfrf: gung lbh bayl *trg* n havdhr fbyhgvba vs gur fhz bs gur svefg guerr qvtvgf vf 8. Guvf vf jung sbeprf gur svefg qvtvg gb or 3 sbe gur rkcrpgrq nafjre frg naq 5 sbe gur bguref.

To put it another way, va lbhe bevtvany cbfg lbh rkcyvpvgyl nffhzr gung gur nafjref gb dhrfgvbaf 2 naq 3 ner obgu "lrf", ohg guvf vf whfg na nffhzcgvba, naq gur snpg gung vg yrnqf gb *n* fbyhgvba qbrfa'g zrna vg yrnqf gb gur *bayl* fbyhgvba. Vg pbhyq pbagvahr vafgrnq:

Yrg'f svk gur frpbaq naq guveq qvtvg naq nffhzr gur nafjre gb dhrfgvba 2 vf "ab" naq dhrfgvba 3 vf "lrf".

Vs gur nafjre gb uvag 5 jnf "ab", jr pbhyqa'g qrgrezvar gur sbhegu qvtvg ng nyy. Gur bayl jnl jr pna or fher bs gur sbhegu qvtvg vf vs gur nafjre gb uvag 5 vf "lrf", naq gur bayl jnl guvf qrgrezvarf n fvatyr qvtvg vf vs gur fhz bs gur svefg guerr qvtvgf vf 8.

Gur svefg qvtvg zhfg or bqq (be dhrfgvba 1 jbhyq or erqhaqnag) naq abg 1 (be dhrfgvba 4 jbhyq or erqhaqnag), fb vg'f rvgure 3, 5 be 7. 7 vf gbb ovt - gur fhz jbhyq or ng yrnfg 10 - fb bayl 3 be 5 ner cbffvoyr. Ohg vs gur svefg qvtvg vf 3, gur frpbaq zhfg or 2, juvpu pbagenqvpgf bhe "ab" nafjre sbe uvag 2. Fb gur svefg qvtvg zhfg or 5 naq gur frpbaq 0.

Fb abj jr'ir svkrq na nafjre frg (ab, ab, lrf, ab, lrf) gung yrnirf hf jvgu rknpgyl bar cva ahzore: svir mreb guerr avar.

I'm not sure I know what you're getting at because:

>But if the first digit is 3, the second must be 2


>So now we've fixed an answer set (no, no, yes, no, yes) that leaves us with exactly one pin number: five zero three nine.

This is not correct, because with (no, no, yes, no, yes) the pin 3039 would also be valid, so five zero three nine is not a unique solution.

Nice puzzle! I am curious about the “there are no redundant questions” part, I don’t think the puzzle would become logically ambiguous or unsolvable if that information wasn’t given, but am I missing something?


Xabjvat gung gurer'f ab erqhaqnag dhrfgvbaf yrgf lbh nccyl fbzr vagrerfgvat ybtvp jvgu ubj dhrfgvbaf 1 & 4 nssrpg rnpu bgure - vg'f abg erdhverq gb fbyir ohg vg'f zber fngvfslvat.

Originally I wanted the puzzle to be heavier on that kind of logic, but I couldn't find a set of questions that worked well.

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