National Cipher Challenge 2019 › Forums › National Cipher Challenge 2019: Countdown to Catastrophe › Free Time Cipher Challenge
 This topic has 11 replies, 5 voices, and was last updated 1 week, 4 days ago by Theletterwriggler.

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21st October 2019 at 9:07 pm #43720TheletterwrigglerMember
In your FREE time have a go at this little CipherMath 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 26Team ‘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 nonconsecutive 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. valueRICKY = 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 = 100add 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.
22nd October 2019 at 12:03 am #43758TheletterwrigglerMemberSorry, 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.22nd October 2019 at 12:03 am #43760MadnessMembermad372
23rd October 2019 at 9:52 pm #43763Spice_master25MemberI might have done my maths wrong or forgotten to add something but spi502.
23rd October 2019 at 9:52 pm #43770TheletterwrigglerMemberTeam THE RILY SIDE (why did they use that?) are smiling
with MAD372 you have the correct names
24th October 2019 at 8:44 am #43795TheletterwrigglerMemberSpice master,
Nothing wrong with your maths – with SPI502 you have the correct namesActually this tells me that all the correct letters were found and I am assuming that the anagrams are easy enough to make the names.
24th October 2019 at 8:34 pm #43826Diamondbandit23MemberI am pretty sure I must have done the maths wrong somewhere but I got DIA328. Is that right? Great challenge, by the way!!!
25th October 2019 at 6:17 pm #43847TheletterwrigglerMemberDiamondband
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 QLAND6th November 2019 at 12:08 pm #43934HoixMemberThat 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 ðŸ™‚
6th November 2019 at 8:23 pm #44048TheletterwrigglerMemberDiamondbandit23
If you want to recalculate…
dia352 would have been correct, a difference of 24.6th November 2019 at 8:23 pm #44050TheletterwrigglerMemberHoix
Then I will also presume you have the correct names ðŸ™‚
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.
4th December 2019 at 12:53 pm #44395TheletterwrigglerMemberDid 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: DENISEWorkins:
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 SIDEABCDEFGHIJKLMNOPQRSTUVWXYZ

00000000011111111112222222
12345678901234567890123456SALLY = 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 
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