Skip to main content

Free Time Cipher Challenge

Viewing 12 posts - 1 through 12 (of 12 total)
  • Author
    Posts
  • #43720

    In your FREE time have a go at this little Cipher-Math challenge.
    In no way let it take you away from the main NCC contest.

    DO NOT POST THE NAMES AND SPOIL IT FOR OTHERS
    You can either keep it to yourself delighting in the fact you solved
    it or post using the method given further below.

    A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

    Team ‘THE RILY SIDE’
    Five crypto solvers got themselves deep into cipher and math!
    Using the KEY above can you name the team members?

    What are thier names?
    Why did they use the team name ‘THE RILY SIDE’?

    Team Member 1:
    a] The two letters that are the same multiply together to make 144.
    b] The highest and lowest values multiply together to make 25.
    c] The capital letter has two digits but three symbols in Roman
    numerals. Also in Roman numerals its form has both rotational
    and line symmetry.

    Team Member 2:
    a] The two letters that are the same multiply together to make 25.
    b] The other three letters are consecutive even numbers which
    multiply together to make 5760.

    Team Member 3:
    a] Using only the first 21 letters, two letters when multipied
    together make 100.
    b] Two other letters when multipied together make 88.
    c] The other letter is a square odd number not equal to its own
    square root.

    Team Member 4:
    a] The two letters that are the same multiply together to make 16.
    b] Two other letters multipy together to make 9.
    c] The last letter is equal to twice the highest valued letter so
    far plus the value of the letter which is neither the highest
    nor the lowest.

    Team Member 5:
    a] The two letters that are the same multiply together to make 25.
    b] Two non-consecutive numbers multipy together make 56.
    c] For the last two letters the lower is a square number and the
    higher is worth one more than twice this lower number.

    This form may help your workings:
    <><><><><><><><><><><>
    Team Member 1: ….. Team Member 2: …..
    Team Member 3: ….. Team Member 4: …..
    Team Member 5: ……

    Workings:
    1a: ………., 1b: ………. 1c: ……….
    [????? is a anagram of ?????]

    2a: ………., 2b: ……….
    [????? is a anagram of ?????]

    3a: ………., 3b: ………. 3c: ……….
    [????? is a anagram of ?????]

    4a: ………., 4b: ………. 4c: ……….
    [????? is a anagram of ?????]

    5a: ………., 5b: ………. 5c: ……….
    [?????? is a anagram of ??????]

    A METHOD FOR POSTING
    ——————–
    1st calculate YOUR name signature, the name
    you use for posting to the forum.
    (Harry is used here as an example) use
    first 3 letters of name HARRY = 8+1+18 = 27
    calculate value of each found name as shown
    here using example names, and add YOUR sig. value

    RICKY = 18+09+03+11+25 = 66 +27 = 93
    BARRY = 02+01+18+18+25 = 64 +27 = 91
    SUSAN = 19+21+19+01+14 = 74 +27 = 101
    FREDA = 06+18+05+04+01 = 34 +27 = 61
    JAQUES= 10+01+17+21+05+19 = 73 +27 = 100

    add the totals 93+91+101+61+100 = 446 so you post sig446 (example HAR446)
    this will help ‘prove’ your post.

    Note if your name is a number only, add the 1st 3 digits.

    #43758

    Sorry, a slight correction.
    Team Member 5:
    c] For the last two letters the lower is a odd square number and the
    higher is worth one more than twice this lower number.

    #43760
    Madness
    Member

    mad372

    #43763

    I might have done my maths wrong or forgotten to add something but spi502.

    #43770

    Team THE RILY SIDE (why did they use that?) are smiling

    with MAD372 you have the correct names

    #43795

    Spice master,
    Nothing wrong with your maths – with SPI502 you have the correct names

    Actually this tells me that all the correct letters were found and I am assuming that the anagrams are easy enough to make the names.

    #43826

    I am pretty sure I must have done the maths wrong somewhere but I got DIA328. Is that right? Great challenge, by the way!!!

    #43847

    Diamondband
    with DIA328 you do not have the correct names (or you have miscalculated)
    Well you did say you were ‘pretty sure’ you have done the maths wrong somewhere
    so your post was correct!!!

    Thanks for comment, and for taking up the challenge.

    To check that you have the correct names do a mod 26 on each names value.
    For example using the first name in the sample
    RICKY = 18+09+03+11+25 = 66 (66 mod 26 = 14 = N)
    For the five correct names you should get QLAND

    #43934
    Hoix
    Member

    That was pretty fun – I can’t be bothered to do the signature thing but the mod of the names I got matches QLAND so I’ll presume they’re right 🙂

    #44048

    Diamondbandit23
    If you want to recalculate…
    dia352 would have been correct, a difference of 24.

    #44050

    Hoix
    Then I will also presume you have the correct names 🙂
    As you did the ‘harder’ mod calc I’ll do the easy addition for you,
    hoi314 is what you would get.

    Thanks for your post, I find recreational maths fun too.

    #44395

    Did you find the crypto team members?
    They were:

    Team Member 1: SALLY. Team Member 2: PETER.
    Team Member 3: KEITH. Team Member 4: DAVID.
    Team Member 5: DENISE

    Workins:
    1a: 12 x 12 = LL, 1b: 1 x 25 = AY. 1c: 19(XIX) = S.
    [LLAYS is a anagram of SALLY]

    2a: 5 x 5 = EE, 2b: 16 x 18 x 20 = PRT
    [EEPRT is a anagram of PETER]

    3a: 5 x 20 = ET, 3b: 8 x 11 = HK 3c: 9 = I
    [ETHKI is a anagram of KEITH]

    4a: 4 x 4 = DD, 4b: 1 x 9 = AI 4c: 2 x 9(I) + 4(E) = 22 = V
    [DDAIV is a anagram of DAVID]

    5a: 5 x 5 = EE, 5b: 4 x 14 = DN 5c: 9(I) x 2 + 1 = 19(S) = IS
    [EEDNIS is a anagram of DENISE]

    THE TEAM NAME?
    Remove the first 3 letters from each name and the
    remaining letters anagram to their team name:
    LY ER TH ID ISE – THE RILY SIDE

    ABCDEFGHIJKLMNOPQRSTUVWXYZ
    ||||||||||||||||||||||||||
    00000000011111111112222222
    12345678901234567890123456

    SALLY = 19+01+12+12+25 = 69 69 mod(26)=17= Q
    PETER = 16+05+20+05+18 = 64 64 mod(26)=12= L
    KEITH = 11+05+09+20+08 = 53 53 mod(26)=01= A
    DAVID = 04+01+22+09+04 = 40 40 mod(26)=14= N
    DENISE = 04+05+14+09+19+05 = 56 56 mod(26)=04= D

Viewing 12 posts - 1 through 12 (of 12 total)
  • You must be logged in to reply to this topic.