Challenges from competitors

[Jbrintcrypt]

@The-letter-wriggler, @Harry
What’s that I hear? Another excuse to spread my ludicrous musings? I am happy to oblige. (For those who don’t know, here’s the puzzle description from last time (with a little change to the year): The following puzzles are all substitution ciphers with a twist: the keywords in each case refers to an event from the year listed. For example, if the event was Coronavirus, the alphabet would be ‘cornaviusbdefghjklmpqtwxyz’ and the text would begin ‘2019/2021: Rhlhgctslqm.’. I have an unusual taste in interesting events, so don’t expect anything that you can find by looking up the year, as these are weird or atypical events.)

Oh, and just to make sure I’ve covered everything- A) I solemnly swear by my troth/ upon the river Styx/ by the Flying Spaghetti Monster e.t.c. e.t.c. that- to the best of my knowledge- these should be age appropriate to appear on the forum, so the moderators do not need to check through before releasing it and B) I realised I made a minor confusion in 1968’s problem: the ‘á’ is also an á in the plaintext, although in context it’s part of a place name so that doesn’t really affect the problem much. Similarly, all numbers refer to (nonsubsituted) years.

1816: Limq Ocshtbm. Licciwdhu qnb Wfm il 1812, qnb SPF tbrdtbt qi osdct f limq qi tblbht fufdhpq fh dhvfpdih lmig Umbfq Omdqfdh vdf Rfhftf. Qwi ybfmp dhqi qnb osdct, dq wfp mbfcdpbt qnfq qnby nft gdpqfabhcy obbh osdctdhu dh Rfhftf mfqnbm qnfh qnb SPF fht wbmb limrbt qi fofhtih dq. Ivbm qdgb odqp wbmb rfmmdbt fwfy oy cirfcp shqdc 1842, dh wndrn tdjcigfqp hbuiqdfqbt qi iwh qnb cfht dq wfp osdcq ih. Flqbm psrrbpp qnby mbhfgbt dq Limq Gihquigbmy. Niwbvbm, ihrb ldhdpnbt jmijbmcy dh 1886 dq wfp mbfcdpbt qnfq qnb Rfhftdfhp tdth’q frqsfccy nfvb f mbfpih qi dhvftb fhygimb fht prfvbhubmp, ihrb gimb, pqicb odqp il qnb fofhtihbt osdctdhu. Qnb mbgfdhp fmb hiw jmdvfqb jmijbmqy.

1886: Sga Ktcoijknjan (jn sga Aac Nojsq). Sgoq oq thjus t fjw jusctwap sntposojf sgts jeeunap of Tdqsanptd etccap Aac Inthhofi. T cova aac wjucp ha gufi tenjqq t etftc haswaaf swj gjuqaq tfp kajkca wjucp sny sj inth os tq sgay qtocap ktqs ljn t etqg knoza. Sgoq soda sga kjcoea paeopap sj qsjk sga avafs hy eussofi pjwf sga njka. Sga qkaestsjnq wana tfianap hy sgoq tfp jfa tsstebap sga kjcoea wosg tf udhnacct. Sgoq ctsan pavacjkap ofsj t ctnia nojs wosg njeb sgnjwofi. Sga faxs pty, sga nojsanq snoap sj tqqtucs sga kjcoea qstsojf, tfp sga docostny wana etccap of, ufljnsuftsacy catpofi sj swafsy qox patsgq.

1919: Rso Gjctqqoq Lcjjm. Rsfq polopq rj ti terutccy ptrsop uikcotqtir fiefmoir ljccjwfia lpjg t rwj gfccfji atccji wtvo jl gjctqqoq, gjvfia tr rsfpry lfvo gfcoq kop sjup, fi Hjqrji (UQT). Qovoptc gtrropq ejghfiom sopo, ques tq kjjp qtlory tim wotb gortc. Moqkfro rso lter rstr U.Q. Fimuqrpftc Tcejsjc, wsj wopo socm poqkjiqfhco, wopo twtpo rstr rso rtib wtq cotbfia tim mtgtaom, rsoy mfm ijr rsfib fr wjucm hpotb qj htmcy. Tq gjctqqoq fq t iji-iowrjifti lcufm, wsopo rso vfqejqfry eti vtpy muo rj ljpeo terfia ji fr (yju kpjhthcy bijw thjur fr wfrs rso ejpilcjwop oxkopfgoir), rso wtvo wtq cfbo t gumlcjw tim rsopoljpo oxrpogocy mtiaopjuq- rwoiry jio wopo bfccom tim jio suimpom tim lflry fidupom.

1968: Detpkqk nhs Ellcjnpr. Eb nhek yspj nhs Ycqnh Ebnsjbpnecbpr Dpjny, eb nhs QKP, njesa nc tsn p det bclebpnsa ucj djskeasbgy. Nhek wpk nc lcgo nhs cnhsj gpbaeapnsk, Jeghpja Bexcb pba Hqisjn Hqldhjsy. Hcwsvsj, eb nhs uejkn djskk gcbusjsbgs Detpkqk pba nhs gpldpetb nspl wsjs pjjsknsa ucj nhs nhjss ucrrcwebt gjelsk: aekcjasjry gcbaqgn, aeknqjiebt nhs dspgs, pba (ly dsjkcbpr upvcqjens) Ijebtebt P Det Nc Ghegptc. Eb nhs njepr, en wpk jslpjosa nhpn pb cuuegsj kpea nhpn nhs det hpa kfqsprsa cb nhsl. Nhsjs ek bcn pby djcvsb ebucjlpnecb picqn whpn hpddsbsa nc Detpkqk punsjwpjak. Kelerpj svsbnk ebgrqas nhs 1967 vcnebt eb nhs ncwb cu Degcpzá, eb whegh p ijpba cu uccndcwasj wcb nhs lckn vcnsk, pba Rpnmenpk, Nsxpk, whegh hpa pb srsgnecb ebgrqaebt nhs gpbaeapnsk cu p wccasb lcasr cu p Bpnevs Plsjegpb, p act pba p tcpn (nhs tcpn wcb. Prr kqiksfqsbn lpycjk hpvs issb tcpnk).

1970: Tjnptc Nvbhtgdcp Ulohn Dcedgncm. Dc Ctsnarnj, o ulohn rneoan rnoelng tc mln etokm tx Tjnptc. Mln ldpluoy gdsdkdtc ocg cosy gnedgng mlom mln rnkm uoy mt gdkbtkn tx mln jnaodck, rnxtjn mlny jtmmng, uok gycoadmn ok dm uok ctm bjoemdeoh mt atsn. Mlny mlnc gnmnjadcng mlom 20 ejomnk tx nvbhtkdsnk utqhg rn kqxxdedncm, gnkbdmn rndcp ogsdkng opodckm qkdcp kt aqel ry oc nvbnjm. Octmlnj dkkqn dk mlom mln ejomnk unjn qcgnj mln ulohn dckmnog tx ocphng mt xdjn dm dcmt mln tenoc. Mln nvbhtkdtc dmknhx gdkmjdrqmng o xdcn kbjoy tx rhqrrnj oejtkk mln kbnemomtjk, hng mt mln nvbnjm’k eoj rndcp kfqoklng ry o hojpn bdnen ocg hnxm tmlnj hojpn bdnenk kmdhh jnaodcdcp tc mln rnoel. Cnuk xttmopn aoy rn xtqcg tchdcn. Dc Iqcn 2020, o bojw uok coang oxmnj mln dcedgncm xtj mln qbetadcp 50ml occdsnjkojy.

And on that note, at the time of writing it’s very late and I should be doing something else. Have fun, if you like that sort of thing.

[Jbrintcrypt]

@Harry

But I did send you the plaintexts and alphabets! Oh, never mind, here they are again:

[You did, and I am truly sorry that we mislaid them. Thanks for sending them again. I will edit the previous post to publish the great challenges! Thank you, Harry]

[Mattyrat2027]

@The-letter-wriggler:

MD5 of your cipher (PLAINTEXT+MATTYRAT2027) = f3da286a8d587cbd956abbcc28aa0e08

I didn’t get a match for your MD5 but I know you had it right from your response. I see your point and will bear it in mind if I do any other challenges. My first Cipher Challenge so I’m still getting used to the forums and how it all works.

How can I contact Harry?

[Kford-academy]

[NOTE: @HARRY, PLEASE DELETE ONE OF THE SECTIONS GIVEN LATER ON IN THIS POST. THAT SECTION IS ONE OF THE ONES IN SQUARE BRACKETS AND IT IS CLEARLY MARKED AS THE ONE YOU SHOULD DELETE. SO NO EXCUSES FOR NOT DELETING THAT PART, OK?]

Here is a review of who has answered my KA2020 challenges so far:

[KA2020-1]
(Post #52729)

@Madness, well done! First one in, as usual…

(Post #52763)

@Person314, well done as well! Note that you were interested about solving the cipher without use of the clues, so I have encrypted the below message using the same cipher, alphabet and key as the final answer of my puzzle (i.e. the answer you posted to the forum).

B. FGH L75 KW8B0 HJK8G8WMQ4 (HWX UWAXK EXX4EEE4 RPPIIA), 0W9 JE4 6RYPTVN1 E O4WH8 OG Q2W1 5JUSUAZRV PEC1RX3 – X0XWV E1SP TI ‘J588’. UL1A FB42I 6ZS 0VVC IIUO8X XMYV WV0YDLST2 Q7EY0GI5 AY1 C0W TUDD BS BVLEX 7ZS 0VVC HJ8E8.
L. FTA XY6 XST7UH VIE5BSWMQ4 (HWX K2YDTRC5 4BM0S679I6031 UQ8PS4), VPTYI ZB US 9MVYYS 6J8OZY6 XV6 78 UDA03 WV. RLN 8WD CLY FSUNLB US I3 4F1 5I5W5A 1EBV OZN 0Y0T 7QFIT H87 YQY1 27J 1UD7 YS7RS BT0WVS9 10 1SE ZWSTS TTDMN4T8T!
B. S7V G07 YVZAK HXJZFSWMQ4 (HWX 3WTXUICM IM2LVV), VZD RE9198S A5I380I5 OY HZM EV14L Q0Z6LKQ. S9 TEZXS8 TX14P, RBCMX0T 1C SZ 80M YW4B0 LFTW 6A MI2X7 AMQ8V 7Y 0PV PSXBHKQ. XZ0S IE6 NQDG Y10 09Q2 V2574 EZ 11 CZE 2XC. VYMLW 3T5U Q9JCW099 T6IW7BT8T0 PTR X4AQ HRL0.
M. TVV 4W8 K2BA0L ISW8HWWMQ4 (HWX X2YHIM96 3EVIBW), XYBSA SG0V9ES UZEA ELXYE TAL R5 FIXVN0T 1NRT 3W8LAC WS9VF7QR (U4I O2H C1QSVGW7VDI) K7 CPZ XMAC (L.K. K7OM59SW), JH KA5KW WB US ‘E15F’ V1 G41 CTN1UXT8T. 1U1W 5B COM9MX7YI ALJ TZRTS HJK4Y0WMQ4, Y2W 627 CNQ I6Y DITBSQ BVG2Y4I5 TZ G6TDX 7ZS N3T0LV.

K P3WI RTS 2U XHT3 VJ4W0!

[KA2020-2]
(Post #57281)

@Person314, thank you for your solution. I have checked and confirmed your answer as correct. I will be posting a cipher in the ‘Not Enough Suffering’ topic soon (once I can get the correct functions out of my 2000 edition of Excel), so look out for that!

[Mattyrat2027]

@Madness I’ve made progress with your cipher, but the possible alphabet combinations are 126,886,932,185,884,164,103,433,389,335,161,480,802,865,516,174,545,192,198,801,894,375,214,704,230,400,000,000,000,000, or more combinations than there are atoms in the universe. Please could you point me in the right direction by at least telling me a program I could use to solve the substitution, because all websites I find don’t accept double letters in the alphabet (which you will understand). Also, with base64, the character count increases by something like 33%, so could you tell me if the substitution is a straight letter for letter?

Thanks

[Madness]

@Mattyrat2027, the substitution is a simple substitution using a 64-letter alphabet. You will not find help on the
internet for this one.

I will let you and the others think about it longer before I give any hints.

[The-letter-wriggler]

@Person314,
That was a really cheeky cipher, love it, as simple as it is, it held the best of us back from a solve for months.
Thanks.

[The-letter-wriggler]

[Harry do not delete the bracketed part below, it is a sample]

@Mattyrat2027 For Your Information (or anyone else)
In #52880 above you said ” My first Cipher Challenge so I’m still getting used to the forums and how it all works. How can I contact Harry?”

When posting a cipher you wish others to try to solve give Harry the plaintext and how it is encrypted inside [square brackets] instructing him
to not show that part, like so…

[
HARRY DO NOT POST THIS PART
PlaintextPlaintextPlaintextPlaintextPlaintextPlaintextPlaintext
TYPE: mono-alphabetic substitution
KEY: WHATEVER
]

THEN YOUR CIPHERTEXT GOES HERE

If you really need to, you can contact Harry via email – [email protected]

[Cribbage]

Madness’ Base64 cipher
[Harry, I apologise for the length of this, but there was no quick way to describe how I have got stuck with Madness’ Base 64 cipher.]
Madness, I am really enjoying your cipher, although I am starting to wonder whether I should have stuck to the cartoons … Below I will try to describe how I have got stuck.
I am assuming that the Base 64 encoding is as described in the Base 64 Wikipedia article:

1. Each plain text character is converted to an 8-bit ASCII binary number, and, as the plaintext is all uppercase characters, this means each 8-bit number is in the range of 65-90, i.e. 0100 0001 to 0101 1010.
2. Each group of 3 plaintext characters, comprising 24 bits in total, is converted into four Base 64 characters of 6 bits each (6 bits giving you 2 to the power of 6 options, i.e. 64), and these 6 bits characters go from 000 000 (translates to ‘A’ in Base 64) to 111 111 (translates to ‘/’), with ‘Z’ being ‘011 001’, ‘a’ being ‘011 010’ etc. (see handy table in the Wikipedia article).

Once this has happened a simple substitution cipher is then applied, but applied to all 64-bit characters, rather than the usual 26.
If the above is correct, then I have hit a problem that I do not know how to resolve and is presumably due to some misunderstanding somewhere.

• An interesting thing to do is to break up the ciphertext into blocks of 4 and to see which characters are at the start of that block. Each block of 4 characters in the ciphertext represents 3 in the plaintext, and the 6 bits of the first character in a block of 4 in the ciphertext must, after the substitution key has been applied, become the first 6 characters of a plaintext character. In 8 bits, all plaintext characters start with ‘010’, and the only corresponding Base64 characters that start with ‘010’ are the letters ‘Q-W’ (we can rule out ‘X’,’010 111’, as well because no ASCII A-Z character starts with those 6 digits). In the ciphertext, the only characters that start a block of 4 are W, C, O, V, N, g, 9, which nicely matches the number of characters in the range of Q-W. This seems promising …
• By repeating this process for the characters at the 2nd, 3rd and 4th positions and working out which of the Base64 characters could successfully fill these slots to create 8-bit numbers in the 65-90 range, we can deduce the following (the characters on the left are the ciphertext letters at their respective positions in the blocks of 4, and the ones on the right are the only actual Base64 characters that can fill these positions)
  1. “CNOVWg9”, “QRSTUVW”
  2. “CPVbmsw7”, “EFUVkl01”
  3. “BNOQVefhkmnsz78”, “BFJNRVZdhlptx159”
  4. “CHJKNOPQUVWfglnoqrsu148”, “BCDEFGHIJKLMNOPQRSTUVWXYZa”
• Again, it seems promising that the numbers on the left and right sides broadly correspond with each other.
• But, when I now analyse this information to narrow down the substitution possibilities, I generate a contradiction: the ciphertext ‘V’ is present in every position in the blocks of 4 and the only actual Base64 character that can fit in each position in a block of 4 is also a ‘V’. In other words, ‘V’ is substituted to … ‘V’, i.e. the 6-bit number for a ‘V’ (‘010 101’) is the only 6-bit number that can be placed at any position in the block of 4 and still be compatible with the corresponding three 8-bit binary numbers in the range 65-90 occupying the same position in the texts.
• The ciphertext ‘N’ is present in 3 positions in the block of 4, and the only actual Base 64 characters that are present in all those positions are ‘R’ and ‘V’, so ‘N’ must be substituted to ‘R’ when decoding, as ‘V’ is already taken.
• But ‘O’ is also present in those same 3 positions, and it can neither be ‘R’ or ‘V’, so I am left with no character for ‘O’ (or ‘N’, depending on the order of these last two bullet-points) …

It feels like I am close, although I have been wrong with this impression in the past. Where have I gone wrong?

[Madness]

@Cribbage, that is not how I would do it. All that thinking would hurt my brain (assuming I am not a noncorporeal heuristic
algorithm floating in cyberspace). But if you want to do all that cognitizing, carry on. I think where you have “gone wrong”
is to assume that ALL possibilities are present in the ciphertext.

Do you want a hint?
UNIRLBHRIREGEVRQGBOERNXGURSENPGVBANGRQZBEFRPVCUREGURSNPGBEVMNGVBABSGURPVCUREVFGURFNZRNFGURPVCUREPYBPXNAQGUVFPVCURE

[Madness]

@Cribbage, sorry about my last post. I must have been having a stroke. I’ll think about it some more. The hint is still
good, however.

[Madness]

@Cribbage, maybe I made a mistake, but the plaintext ends with a newline character. So throw out the last block when
you analyze the ciphertext. That might fix your approach, by removing ‘N’ from 3. and ‘1’ from 4..

[Kford-academy]

KA2020-3

Anyone stuck for something to do? Here’s a nice mathematical puzzle!

Find:
1. A four-digit number that is one-ninth of its reverse
2. A four-digit number that is one-quarter of its reverse
3. 9 two-digit numbers that are equal to the sum of their digits plus the product of their digits
4. A two-digit power of 2 that is 5 more than a cube
5. 2 numbers that are one-sixteenth of their cubes
6. A number that is 1.5 less than 1.5 times itself
7. 2 numbers (above 0) that have the same amount of factors as the numbers themselves
8. 2 numbers (below 100) that are one-half of the sum of their factors (including themselves)
9. A number that is equivalent to the infinite series 2/3 + (2/3)^2 + (2/3)^3 + (2/3)^4 + …

Post your answers in the form of a matrix, multiplied by another matrix (obviously post the answer to the multiplication as well). Also post what you have put in each matrix. (Note: when posting matrixes, post them in rows, from left to right and going from top to bottom. Use semi-colons to separate rows and spaces or commas to separate numbers within the same row. For instance, [1, 2, 3; 4, 5, 6; 7, 8, 9] and [1 2 3; 4 5 6; 7 8 9] both correspond to the following matrix:
|1 2 3|
|4 5 6|
|7 8 9|
) For instance, if I thought the answer to question 1 was 1, the answer to question 2 was 2, one of the answers to question 3 was 3, and so on, I could post my answer in the following form:

[ANSWER1, ANSWER2, ANSWER3; ANSWER4, ANSWER5, ANSWER6; ANSWER7, ANSWER8, ANSWER9] * [1; 2; 3] = [14, 32, 50]

However, if I thought all 9 answers to Question 3 were 1, 2, 3, 4, 5, 6, 7, 8 and 9 and I only wanted to submit those, I could post the following:

[ANSWER3 (lowest), ANSWER3 (2nd lowest), ANSWER3 (3rd lowest); ANSWER3 (4th lowest), ANSWER3 (5th lowest), ANSWER3 (6th lowest); ANSWER3 (7th lowest), ANSWER3 (8th lowest), ANSWER3 (highest)] * [1; 2; 3] = [14, 32, 50]

Ideally, the second matrix should have more columns than rows, so that there will be multiple correct answers that could have been in the first matrix (making it harder for anyone to get the answers off someone else).

[Mattyrat2027]

@Madness could you give one tiny hint as to whether padding has been used in the base64, before the substitution? I notice that = has not been used in the substitution, but then again = is not a base64 character. Is this what you were hinting at when you said

“@Cribbage, maybe I made a mistake, but the plaintext ends with a newline character. So throw out the last block when
you analyze the ciphertext. That might fix your approach, by removing ‘N’ from 3. and ‘1’ from 4..”

and we have to start analysis without the last block?

Thanks

[The-letter-wriggler]

@Jbrintcrypt
Well that was a little bit of history I would not have read otherwise.
What a crazy world we live in eh!
Thanks.

Do you want people to post MD5 proves?

Here I have done MD5’s on the keys in uppercase+TLW on the end.

For instance 2019/2020 CORNAVIUSBDEFGHJKLMPQTWXYZ + TLW
so CORNAVIUSBDEFGHJKLMPQTWXYZTLW
gives MD5: 1b54e9a894f0dd935e53ee0358d92e0d

1816 MD5: 39bf1774a36fa042352fae2ce401bbca
1886 MD5: 20e5649880b11efbc980889a910cf04f
1919 MD5: 61e7099c1b1c9a8880cd3c6975ddf03f
1968 MD5: 116dbaa17d88b71317858554a40875e3
1970 MD5: 4cc7dec6c84006e2748e3f9a8d54f97b

[Author]

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