Ciphers and Puzzles from Competitors

[10degrees-admin]

KA-1 Hint One!
120102412120 40241012 141000242134? 1022 1201, 111221 13 12103110021334 141000242134 4001 1201024121 402421 0134102310321302 013221, 041140 1032 4024342121 20103121321210013212…

[10degrees-admin]

Is there any interest is trying some stream ciphers?
If not, I will save it/them for another season.

TLW A Six Fingered Alien.

Calling all mathematicians (that’s all of you right?)

Can any one verify my answer?

In 2001 Clifford A Pickover posted ‘Grand Internet Math Challenge’ where he gave 23 challenges.
One of those was the following:

#3. A Six Fingered Alien:
A six fingered alien comes to you and asks, “What is the sum of all the numbers that can be formed with the digits 1,2,3,4,5, and 6, using each digit twice in every number.”
What is your answer? How did you solve this?

So using the 12 digits 123456123456 add up all the >unique< numbers that can be formed from them.
What is your answer and how did you work it out?

I did this and my answer is 2910599999997089400 but I do not know if it is correct as I never saw any answers or workings for it. I can give my workings but I will withhold them for the obvious reason.

So is 2910599999997089400 correct?

[Nice challenge! Harry]

[10degrees-admin]

Yeah, I’d be interested in a stream cipher Madness, I’ve never looked at them before so I’d be interested to see one

TLW CIPHER Hint Posting 5 FINAL

Hopefully you should be able to solve it with this.
The hint is slightly cryptic too.

FUUJFWX
HMTTXJ
INANXTW
MNLM
QJYYJWX
QTB
UFNWX
UQFNSYJCY
WJAJWXJ
WNLMY
XUQNY
YMJ
YBJSYDKTZW
ZSYNQ
ZXJ
HJ YJ WY YW QX ZJ QB MM IW UX XY WJ ZQ UY FX

[10degrees-admin]

@Kford-academy KA-1 SLE CPE. Quite glad you didn’t add the fractionation step in this cipher.

@The-letter-wriggler I got the same answer for your six fingered alien puzzle.

[10degrees-admin]

@67105112104101114 “CIPHER”

OK, so there are two types of stream cipher. Synchronous stream ciphers
create a key stream that does not depend on the plaintext. A very simple
example is the Trithemius cipher. Here, the key stream is 0, 1, 2, 3, …
The ciphertext is formed by adding numbers from the key stream to the
plaintext modulo 26. E.g.,

T H I S I S T H E P L A I N T E X T (plaintext)
19 07 08 18 08 18 19 07 04 15 11 00 08 13 19 04 23 19 (plaintext values)
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 (key stream)
19 08 10 21 12 23 25 14 12 24 21 11 20 00 07 19 13 10 (ciphertext values)
T I K V M X Z O M Y V L U A H T N K (ciphertext)

If you want to see a very complicated synchronous stream cipher, take a
look at the Solitaire cipher at https://www.schneier.com/academic/solitaire/

An asynchronous stream cipher uses the plaintext in making the key stream.
I.e., the values of the current plaintext character and any previous
characters can enter into the calculation of the current number in the
key stream.

Here is a ciphertext made from a very simple asynchronous stream cipher:

YMIBHVZRGZUHTLPKZHGLTNQRBNNQBMIEXBKEBFDGGFGKWVJEAINAVDTRRHDNQFNZGWHDCQIJAPS
TYBUQWODPTPLPAEUVQZWSXTFRMAAQRGYNTWPHZEBAPVRBMATKCBMPFVNMXVIZJMJYTMKRCVBDZZ
DAAWBFGTFRKABLUHRGLZVIMWFFMVTIQBFMGQHZEBAMNZVENQDRKABLYPUEWPQWLHAVVTYUTVEMU
VTLWFAFALFMLZHVTXLCVJEBAHTEPESLMAYVILLJYRQHTERJHGFHKLXLVTLWFATPATBCMWDGGFGK
WVNWSGTYXROPHALVVHZEBACDAAWBFDTSOXALWWENQWALCKMDHLPURJIPGZVPLRVMJIMZLWLNNMX
VVHITFRMAAQRGTNQWALRCJFWPHKPUEWPQYHUKXALVDDUHRGALWSILKLPWZMCWKSILDRKABLJGAQ
JFCVNULTVSTFDMTBWBSTFPMIBHVZCMIUNLGLHEHVKIGWJKDMMGKXAHTUVFWORNDVNKIFOVMZKJV
ZNNRDMLEPXBBKFZWDLKRIONAHJZVPMIBVKIGWJZSILDRKABLEBBRETTLWDSILPDGGFGKWV

And here is a slightly more complicated one:

ZZOVJMEGXBBNRUUNVOHVNUOHBQRLELTLGUCEIEEWHVNGOBUBFIQDRWSZZSLSWUKEMFJWWPJAALW
UUMEYKORNNFFUJAOOQXFSYBFXQHBDWESFFSVYMZSTXXCTTBESXQXBEEVFJMPXBXXFQUXQXBTAIX
FSXIIBUYLTGMAUNUUXFSXXZSXIIBUYLPSGATNCUCFJMAWJJWZKIQDJXKDKOQUCNVIOOGZGKIEIZ
DLEAASFTPIWPPALJRDSGYQYZKOTHYYLJXCVCGIZDZSGXBBDKDKOUOWZZMOSKIATXJBBUNUYCPSG
LELPHAOFRTKYMZDGKNQYCVCGZVJVDVNVGKCNBPBFIUUMEMHLWUIVOVZRTKOSFXXPIPTRKRLYBFW
ADWKGGXASLSWOVDSASSYMXAERJTRSMFGEERVSLCCQHKSFFWEPNYSUECELLYAEXEIGNNQDRKIMFK
EPAYAOFWACVZCVCGOFKVDJQJYYRYQJXQXXQEJCJNRIZZSACCNYWSWWRZMSKRZOOBEXEIGVVNFJM
DLRYRLYBFWEXXKNGNRJFJNGYGRMQHZNAGYMRKRVGGXH

Neither one involves anything complicated, so don’t over-think it.
Good luck.

[10degrees-admin]

KA-1 News:
Bubble-sort has the first correct answer! I now await other correct submissions.

[10degrees-admin]

KA-1 Hint Two!
CQTX BRV HZAIVVFVW OPK SEMXMIIR GVTYIM. IRP BXYIM PORGW GVJDOHR LZROA UR BXYIM PORGW. YMIB ZLEIV ADTR FR E KVVVYTBWZXDWT GVTYIM.

[10degrees-admin]

@TLW

Good news and bad news. The good news is that I can replicate your result for the
six-fingered alien. I did it by looping over all of the integers from 0 to 12!-1,
and converting each to a factoradic number (google me!), which is easily converted
to a permutation of 0,…,11. I rejected any permutation in which x+6 came before x,
for any x in 0,..,5, in order to eliminate duplicates. Then each permutation was
converted to a permutation of 1,…,6 with two copies of each digit. The permutations
were converted to integers and added to a running sum.

The bad news is that it is wrong. Here’s why:

How many fingers does a six-fingered alien have? You might say six: three on each
hand (I’m assuming it has two hands). But then it would likely use a base-6 number
system. But “6” is not a digit in base 6, any more than “ten” is a digit in base 10.
So probably the alien has six fingers on each hand. This makes sense, because we
say that the three-toed sloth has three toes because it has three toes on each
foot, rather than three toes distributed over all feet. OK, so the alien probably
uses a base-12 number system. Your calculation assumed that all the numbers were
in base 10. If instead, we evaluate each of its 12-digit numbers as base-12 numbers,
then the answer (in base 10) is

21232801607474457000

and in base 12 (new digits are “a” and “b”, but we could use dozenal notation
instead and use “T” and “E” or similar) is

96a15bbbbbbb251a60

Now, here’s something interesting: The sum of the digits 1,…,6 is 21. And
the sum of such digits in the twelve spots is

2100000000000 + 210000000000 + 21000000000 + 2100000000 + 210000000 + 21000000
+ 2100000 + 210000 + 21000 + 2100 + 210 + 21 = 2333333333331.

This wondrous number evenly divides the sum 21232801607474457000, and
21232801607474457000 / 2333333333331 = 1247400

This last number, 1247400, represents the many ways of assigning the remaining
11 digits to their positions once we have fixed one. It is 11! / 2^5, since
there are 11 things to place in 11 spots, but five pairs of identical items
that we do not want to double-count.

What if we do this again in base 12? Then the sum of the digits 1,…,6 is 19,
and the sum of putting them in each of the twelve spots is

1900000000000 + 190000000000 + 19000000000 + 1900000000 + 190000000 +
19000000 + 1900000 + 190000 + 19000 + 1900 + 190 + 19 = 1aaaaaaaaaaa9

Now 96a15bbbbbbb251a60 / 1aaaaaaaaaaa9 = 501a60, but this is again 1247400 in base 10.

TLW A Six Fingered Alien.

@Bubble-sort Thanks for your confirmation. (Now we cannot both be wrong – can we?)

Prove you can do it…
I’m accepting as many confirmations as possible, so if anyone is able to work it out please let me know.
I may ask you for your workings though, so do not post if you are unable to supply them should I ask.

[10degrees-admin]

Some mono sub ciphers I came across while websurfing:
https://qmplus.qmul.ac.uk/pluginfile.php/1893653/mod_resource/content/3/AllSubCips2019.txt
Might make good practice.

@Madness That’s a nice find, 77 ciphers to while away the time.

Re: Mono Ciphers.
Just realized that these are from [whoops, accidentally deleted this. Harry].
I have hundreds of types of ciphers from there, many I have not yet solved.
They do an in house cipher exam each year where the students make and submit ciphers and have to solve each others for exam points.
Thought you’d like to know.

[Hmmmm, not sure I should be approving this. It is a fine institution, but they can do their own advertising. Harry]

TLW A Six Fingered Alien.

@Madness. A fascinating read, you have a wonderful mind!

I do not know if you are over estimating the problem or if I have under estimated it.

I have worked it out logically using ordinary base 10 maths.

We need more input on this. Any one else got ideas?

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