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The National Cipher Challenge

Puzzles

Viewing 15 posts - 31 through 45 (of 47 total)
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  • #97713
    robb
    Participant

    @Kingswinford, so as not to give too much away, I might have included a couple of words in my response that show this… if Harry allows …

    Harry, If it is ok (?) then Puzzle 1 is MD5(“SOMETOUGHPUZZLES”) and Puzzle 2 is MD5(“WILLYOUSOLVETHEM”). These both give the correct MD5 values.

    #97443
    CaldewBA-Alfie
    Participant

    Here’s my solution:
    Cut the cake into quarters looking down on it to create 4 slices. Then cut horizontally through the cake to half them all and create 8 slices.

    #98090
    RickOShea
    Participant

    In conversation Ms Warne recounted that she had overheard the crew discussing Pacific’s progress across the Atlantic Ocean. With favourable westerly winds Pacific was able to make 12.5 knots on the journey from west to east, but battling the same winds, could only manage 10 knots on the return leg, from east to west. So, it would take two and a half days longer to complete the return leg of the journey.

    How far was Ms Warne’s trip in nautical miles?

    One knot is equivalent to one nautical mile per hour. The crew were experienced sailors and reckoned all times in Greenwich Mean Time avoiding any complication associated with time zones.

    #98150
    Kingswinford-Warriors-Alumni
    Participant

    @robb, #97713
    I can officially say that you are correct! (Cue applause.) The challenge for everyone else now is to work out how those answers are obtained…

    I am still working on my final two puzzles (half terms are always busy!), but hope to post them shortly. The next puzzle should be easier!

    #98141
    robb
    Participant

    @RickOShea, nice little algebra problem.

    Lets call the distance travelled in one direction d.
    We also know that distance = speed x time and we are working in hours as knots = miles per hour

    For journey 1, then d = 12.5 * t_1
    For journey 2, then d = 10 * (t_1 + 2.5*24) = 10 * (t_1 +60) = 10*t_1 + 600 Note: two and a half days longer than journey 1 and there are 24 hours in a day

    As the distance d is the same for both we can write that 12.5*t_1 = 10*t_1 + 600
    Solving for t_1 gives (12.5-10) * t_1 = 600, t_1 = 600 / 2.5, therefore t_1 = 240 hours, or 10 days.

    Journey 1 therefore was 12.5 * 240 = 3000 miles in 10 days
    Checking, journey 2 was 10 * (240+60) = 3000 miles (as we would expect) in 12.5 days, or 2.5 days longer than journey 1.

    #98173
    ByteInBits
    Participant

    Soooo, no one has any thoughts about post #97973 this thread?

    Has it defeated everyone?

    #98194
    October__Rain
    Participant

    To find the distance of Ms. Warne’s trip in nautical miles, we can use the information provided about the speeds and the time difference between the two legs of the journey.

    First, let’s denote:
    ( D ) as the distance of one leg of the journey in nautical miles.
    ( V_1 = 12.5 ) knots as the speed from west to east.
    ( V_2 = 10 ) knots as the speed from east to west.
    The time taken for the west to east journey is ( \frac{D}{V_1} = \frac{D}{12.5} ) hours.
    The time taken for the east to west journey is ( \frac{D}{V_2} = \frac{D}{10} ) hours.
    We know the east to west journey takes two and a half days longer than the west to east journey. In hours, this is ( 2.5 \times 24 = 60 ) hours.
    So, we have the equation:
    \frac{D}{10} = \frac{D}{12.5} + 60
    To solve for ( D ), first eliminate the fractions by multiplying the entire equation by 125:
    12.5D = 10D + 750
    Simplifying gives:
    2.5D = 750
    Solving for ( D ):
    D = \frac{750}{2.5} = 300
    Thus, the distance of Ms. Warne’s trip in nautical miles is 300 nautical miles for one leg of the journey, making the round trip 600 nautical miles.

    WHERE DID I GO WRONG @robb?

    #98231
    Kingswinford-Warriors-Alumni
    Participant

    Here is an answer and a puzzle…

    ——

    @molleculeYT, #96310

    (I don’t know how it took me this long to get to this puzzle, but…)
    If we need to cut the cake into 8 equal sectors, first divide it into quarters as normal. Then rearrange the four pieces so that their planes of symmetry are all aligned. You can now cut along this plane to get 8 identical pieces!

    ——

    And now a new conundrum! (This one is definitely easier than the ones I have done thus far.)

    Puzzle Hunt, Puzzle 4: Another Answer (1,6,9)
    MD5: 32afd3043cc56dd86fa31f6d711f40b3

    [PUZZLE]

    I hope you’re not on the fence about solving this puzzle; this one is off the rails!

    MRSLE YISIT EETRF THTA
    MAWOO XYSSD IRRER EASIS ETAVR HRMED OTHSI EIOPC NCACE GNLFL OIAEL E
    MSSOS NYIAA KSAII NETDH TALTV OALHR ADNAE ENHIR DLC
    MEMIT DYIYL IOPTO ASLBR FLAWO SCEIN HHUEE AIRHA TTOVM

    [N.B. Please decrypt each line separately. This note is not a puzzle.]

    [SOLUTION]

    As hinted at by the introduction, each line was encrypted with the rail fence cipher (but with different keys each time). The decrypted messages are:

    (Key: 3) MY FIRST IS THE LETTER A
    (Key: 6) MY SECOND IS ASSOCIATED WITH CAESAR OR VIGENERE OR HILL, FOR EXAMPLE
    (Key: 5) MY THIRD IS A HARD TASK, AND ALSO A TELEVISION CHANNEL
    (Key: 4) MY WHOLE IS A COMPETITION THAT SHOULD BE VERY FAMILIAR

    These clues lead to the final answer: A CIPHER CHALLENGE

    #98328
    Kingswinford-Warriors-Alumni
    Participant

    #98231 (Puzzle Hunt, Puzzle 4)

    Well that didn’t go quite as planned, did it? I suppose I’ll have to render this puzzle as void now (for the purposes of the Puzzle Hunt) – but this does mean I can create a new puzzle in its place! Stay tuned…

    #98396
    robb
    Participant

    @October__Rain #98194
    Your multiplication of 60 is incorrect: 60 * 125 = 7500 and not 750. Easily done.

    #98397
    robb
    Participant

    @ByteInBits Original #97973 & follow-up #98173

    Don’t be bridled about it, maybe we were leading you on 😉

    The stone is “TO TIE HORSES TO”.

    #98440
    LH24
    Participant

    very good- had to think. The single layer clue helped- wouldn’t want to miss out on any of the filling.

    #98483
    October__Rain
    Participant

    Thanks Robb. Can’t believe I didn’t notice that lol

    #98514
    RickOShea
    Participant

    5827b072cc2p8190772g6710c83f581f2v1v3w1l758164mb197z19353f34906d581658191m02626gm375f91q4165ly471dl1551970910772g61d58191m02626g64mb197f2v1536261.0ll7p0c1709191m3v1d.0915819106yc2p8197f93m581p91v2347y109’7f394m7w82c880v1635b116b93k16?

    Not sure if any of you will get this, but give it a go! Harry

    #98539
    Kingswinford-Warriors-Alumni
    Participant

    Here is a new puzzle! Hints will be available on demand.

    Puzzle Hunt, Puzzle 4.5: Annoying Answers (12,4)
    MD5: 52d7471be9f6359e9bd0ed82b89511cd

    [PUZZLE]

    Scouring over my personal archives, I found this completely nonsensical sentence, together with a weird grid and the message “What vowel-y clue lurks within these words?”. Someone help me!

    MESSAGE START
    
    It’s unusual to say that reloaded colours burst into fantasy. Only sums can join that beauty - it’s weird!
    
      |  2
      |aeiou
    --+-----
     a|POLYB
     e|IUSTV
    1i|WXZAC
     o|DEFGH
     u|KMNQR
    
    MESSAGE END
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