Ciphers and Puzzles from Competitors

[10degrees-admin]

@Snakecharmer314,
The only thing that could make sense, if you are not just making random strings, is some sort of base 11 to 26 conversion. If this is the case, just say so, but don’t give any more information, so I can work out the details for myself.

Re: TLW TRIPLET TRIP
Forgot to state:
Moves are orthogonal – left-right-up-down within the grid (no wrap-around).

[10degrees-admin]

TLW THE HUTCHINSON PROBLEM
I have an 8-move way:
2D 3D 3R 3U 1R 3D 2L 3R

Re: TLW THE HUTCHINSON PROBLEM
@Kford-academy I have a 6 move! Care to find it?

@Madness they are not just random strings.

[10degrees-admin]

Madness: I can verify that there is meaningful English encoded in Snakecharmer314’s ciphertext.

I’m not sure what you mean when you say the “ciphertext has maximal entropy no matter how it’s sliced, and therefore is also random”, but I’m pretty sure you’re not using the word “entropy” correctly. Can you explain what you think you’re saying?

In any case, it’s impossible to say anything about the entropy of a single block of text if you don’t know how it’s generated. For an analogy, suppose I have some random event, and I tell you the outcome was “4”. What is the entropy? It depends on how I generated the outcome.

If the outcome was determined by rolling an ordinary 6-sided dice, then the process has a binary entropy of log_2(6)=~2.585 (the base 2 logarithm of 6). If I instead rolled an octahedral dice, then the binary entropy is 3. And if I used the process described in xkcd.com/221, then the entropy is 0.

[10degrees-admin]

Kford-academy: In post #49107 your equation is missing a factor of two – it should be x=gt^2.
You can work this out or check it using “distance = average speed x time” and “average speed = 1/2 x accelleration x time” (assuming accelleration is constant, which it is in this case).

TLW ANSWERS TO MY POST – PART 1
===============================

============================================ ANSWER
TLW CIPHER
==============================
Admittedly this seems a hard one to crack.

I came across a similar looking cipher on subRedit and came up with this
idea to try to crack it but it failed to solve it.

Then I thought I would use my idea to make my own cipher.
So that is how I came to make this one.

Maybe it’s possible to solve as a large anagram, but I think that is a big ask.

ALL THE HINTS BROUGHT TOGETHER AND GIVEN IN SOLVED PLAINTEXT
=============================================================
1] cipher letters are usually given in this type of format. (UPPER CASE)
2] Your hint is the shift of this cipher also read the very puzzle/cipher itself for another! (24, HINTING AT THE NUMBER OF LETTERS, THEY NEED TWISTING IN LINES)
3] one particular math op will help you. (DIVISORS, THE TWISTING PARTITIONS)
4] use divisors wisely then divide and conquer each with a twist. (PAIR THEM)
5] choose the right twentyfour letters, use low high divisor pairs, reverse, split, until plaintext appears (ALMOST THE SOLUTION)

THE CIPHER:
===========
Yet To Solve ThiS One
tHe hIntS
eVery LIne Will EacH
ONLy BE a LittLE Twisted

THE PLAINTEXT:
===============
ONLY THE BEST WILL SOLVE THIS

THE SOLUTION:
=============
Use only the capital letters (the lowcase letters are dressing used as a hint)
Count of the capital letters = 24

Divisors of 24 = 1, 2, 3, 4, 6, 8, 12, 24.

Pair the low-high and high-low divisors:

1:24, 2:12, 3:8, 4:6, 6:4, 8:3, 12:2, 24:1

(for this cipher the 12:2 pair were not used but could have been)

Here are the steps to a solve:
01 Remove all lowercase letters
02 Make one group of the 24 capitals keeping them in the order given (1:24 pair)
03 Reverse the group (the twist)
04 Split to make 2 groups of 12 letters (2:12 pair)
05 Reverse each group
06 Split to make 3 groups of 8 letters (3:8 pair)
07 Reverse each group
08 Split to make 4 groups of 6 letters (4:6 pair)
09 Reverse each group
10 Split to make 6 groups of 4 letters (6:4 pair)
11 Reverse each group
12 Split to make 8 groups of 3 letters (8:3 pair)
13 Reverse each group
14 Make one group of the letters and edit to plaintext (24:1 pair)
===================================================================
Workings:
=========
01+02 = YTSTSOHISVLIWEHONLBELLET
03 = TELLEBLNOHEWILVSIHOSTSTY
04+05 = WEHONLBELLET YTSTSOHISVLI
06+07 = EBLNOHEW TSTYTELL ILVSIHOS
08+09 = HONLBE YTSTWE LILLET SOHISV
10+11 = LNOH TYEB EWTS LLIL OSTE VSIH
12+13 = ONL YTH EBE STW ILL SOL VET HIS
14 = ONLY THE BEST WILL SOLVE THIS

———- HOW TO MAKE THIS TYPE OF CIPHER ————

Encryption steps:

ONLY THE BEST WILL SOLVE THIS — Choose a 24 letter phrase

Note the use of divisors of 24 : 1, 2, 3, 4, 6, 8, 12, 24.

ONLYTHEBESTWILLSOLVETHIS — 1×24 removed spaces
SIHTEVLOSLLIWTSEBEHTYLNO — reverse group
SI HT EV LO SL LI WT SE BE HT YL NO — set 12×2 (extra step to the one I gave)
IS TH VE OL LS IL TW ES EB TH LY ON — reverse each group
IST HVE OLL SIL TWE SEB THL YON — set 8×3
TSI EVH LLO LIS EWT BES LHT NOY — reverse each group
TSIE VHLL OLIS EWTB ESLH TNOY — set 6×4
EIST LLHV SILO BTWE HLSE YONT — reverse each group
EISTLL HVSILO BTWEHL SEYONT — set 4×6
LLTSIE OLISVH LHEWTB TNOYES — reverse each group
LLTSIEOL ISVHLHEW TBTNOYES — set 3×8
LOEISTLL WEHLHVSI SEYONTBT — reverse each group
LOEISTLLWEHL HVSISEYONTBT — set 2×12
LHEWLLTSIEOL TBTNOYESISVH — reverse each group
LHEWLLTSIEOLTBTNOYESISVH — the 24 letters to use or…
HVSISEYONTBTLOEISTLLWEHL — …reverse group and use these

Now make up a meaningfull sentence using lowcase letters…
HVSISEYONTBTLOEISTLLWEHL — must have at least one capital in each word
————————– — to make the Cipher
HaVing Some ISsues
thEY ONly Take BiTs
LOok ovEr thIS
Twisting weLL
With Each Half Line
————————–
or without reversing in the last step…

LHEWLLTSIEOLTBTNOYESISVH — capitals to use in your words
————————– — to make the Cipher
Let tHE WiLL To Solve thIs
bE OnLy To Beat iT NOw
YES It iS Very Hard
————————–

============================== EXAMPLE 2 new plaintext, new cipher

SO VERY HAPPY THAT I SOLVED IT

Encryption steps:

SOVERYHAPPYT HATISOLVEDIT
TYPPAHYREVOS TIDEVLOSITAH
TYPPAHYR EVOSTIDE VLOSITAH
RYHAPPYT EDITSOVE HATISOLV
RYHAPP YTEDIT SOVEHA TISOLV
PPAHYR TIDETY AHEVOS VLOSIT
PPAH YRTI DETY AHEV OSVL OSIT
HAPP ITRY YTED VEHA LVSO TISO
HAP PIT RYY TED VEH ALV SOT ISO
PAH TIP YYR DET HEV VLA TOS OSI
PA HT IP YY RD ET HE VV LA TO SO SI
AP TH PI YY DR TE EH VV AL OT OS IS

Make up a meaningfull sentence using lowcase letters
APTHPIYYDRTEEHVVALOTOSIS — use the capitals in each word
————————– — to make the Cipher

A ParTial Hint
Put In equallY evenlY anD Reverse ThEm gEt tHe
reViVAL Of The Original Script InStantly

————————–
Decryption steps:

APTHPIYYDRTEEHVVALOTOSIS — The capital letters only

AP TH PI YY DR TE EH VV AL OT OS IS — set 12×2
PA HT IP YY RD ET HE VV LA TO SO SI — reverse each set
PAH TIP YYR DET HEV VLA TOS OSI — set 8×3
HAP PIT RYY TED VEH ALV SOT ISO — reverse each set
HAPP ITRY YTED VEHA LVSO TISO — set 6×4
PPAH YRTI DETY AHEV OSVL OSIT — reverse each set
PPAHYR TIDETY AHEVOS VLOSIT — set 4×6
RYHAPP YTEDIT SOVEHA TISOLV — reverse each set
RYHAPPYT EDITSOVE HATISOLV — set 3×8
TYPPAHYR EVOSTIDE VLOSITAH — reverse each set
TYPPAHYREVOS TIDEVLOSITAH — set 2×12
SOVERYHAPPYT HATISOLVEDIT — reverse each set
SOVERYHAPPYTHATISOLVEDIT — set 1×24
SO VERY HAPPY THAT I SOLVED IT — edit to plaintext

====================================================================
============================================ ANSWER
TLW CIPHER #2
==================================
You may have to get on the telephone to tri to solve this cipher!

THE SOLUTION:
=============
TYPE : Telephone keypad cipher with three alterations.
KEY :
2 3 4 5 6 7 8 9 1 0
====================
A D G J M P T W Q sp
B E H K N R U X Z *
C F I L O S V Y * *

Used button 0 for space.
Moved q from button 7, and z from button 9 and put them on button 1.
* are spaces only used as a place holder in the 2nd & 3rd rows.

THE FULL PLAINTEXT:
====================
this message was right under your fingertips on a standard keypad telephone the ten digits used are zero to nine with digit zero representing a space and digit one representing q or z each of the other digits two to nine represent one of three possible letters these you will find on the numeric buttons of a phones keypad

WORKINGS:
==========
Here is the decrypt kept to 60 characters per line.
It is now up to a human to make each meaningful word by
taking one letter from one of the three (tri) lines.

THE SOLVE (each number has 3 letters):
844706377243092707444808633709687034643784770660207826327305
============================================================
tGGP mDPPagD waP PGgGt TMdDP WMTP DGMgDPtGpP MM a PtaMdaPd J
UhHR NeRRBHe XBR rHHhU unEer XNur EHnHerUHRR Nn B RUBnEBrE k
VIis OFssCIF YCs SiIIV VOFFS yoVS fiOIFSViSs oO C sVCOFCSF L
============================================================
this message was right under your fingertips on a standard k

397230835374663084308360344487087330273013760860646309484034
============================================================
DWpad tDJDpGMMD tGD tDM dGgGtP TPDd aPD QDPM tM MGMD wGtG dG
eXRBE UeKeRhNne Uhe Uen EHHHUR uReE Bre zerN UN nHne XHUh EH
FySCF VFlFSIoOF VIF VFO FiIiVs VsFF CSF FSo Vo OiOF YiVI Fi
============================================================
eypad telephone the ten digits used are zero to nine with di

448013760737737368464020772230263034448066307377373684640106
============================================================
gGt QDPM PDpPDPDMtGMg a PpAAD aMd dGgGt MMD PDpPDPDMtGMg q M
HHU zerN reRreRenUHnH B RRBBe BnE EHHHU Nne reRreRenUHnH Z N
IiV FSo SFSSFsFOViOI C sScCF COF FiIiV oOF SFSSFsFOViOI o
============================================================
git zero representing a space and digit one representing q o

701032240630843068437034448708960860646307377373680663063084
============================================================
P Q DaAG MD tGD MtGDP dGgGtP twM tM MGMD PDpPDPDMt MMD MD tG
r z eBBh NE Uhe NUher EHHHUR UXN UN nHne reRreRenU Nne NE Uh
S FCcI of VIF oVIFS FiIiVs VYo Vo OiOF SFSSFsFOV oOF of VI
============================================================
r z each of the other digits two to nine represent one of th

733076774253053883770843730968094550346306608430686374202888
============================================================
PDD pMPPGAJD JDttDPP tGDPD WMT wGJJ DGMd MM tGD MTmDPGA ATtt
ree RNRRHbKe KeUUerR UheRe XNu XHKK EHnE Nn Uhe nuNerHB buUU
SFF SossiClF lFVVFSs VIFsF yoV Yill fiOF oO VIF OVOFSic CVVV
============================================================
ree possible letters these you will find on the numeric butt

6670630207466370539723
======================
MMP MD a pGMMDP JDWpad
NnR NE B RhNneR keXRBE
oOs of C SIoOFs LFySCF
======================
ons of a phones keypad

This cipher could be better made using double digits.
Instead of 8447 we use 81 42 43 73 where the second
digit is the row reference making for auto decoding.

====================================================================
============================================ ANSWER
TLW MINI CHALLENGE ALL THE SOLUTIONS
====================================

THE PLAINTEXTS:
================
CIPHER ONE
==========
you have clues and anagrams to help you get started with each cipher.
for instance the following should help you open up the next cipher, number two…
loves to rearrange the message in a way that makes it unreadable, but the clue here will help!

CIPHER TWO
==========
welcome to the world of ciphers – you have just decrypted the second cipher. ‘loves to rearrange’ was obviously the clue, an anagram of ‘loves’ giving the word ‘solve’ which is what you did! will there be a clue to the following cipher within the one just solved? re: leaving together this clue will help you see the next cipher that follows.

CIPHER THREE
============
its looking good now that the second cipher is cracked, ‘re: leaving together’ was the clue so ‘re’ and ‘leaving’ together anagram to revealing, you are surely on your way to fame so keep at it. the next challenge is just that, something you will have to queue for, but with cunning, insight and perseverance you will get your reward.
(Note I should have said ‘third’ not ‘second’)
CIPHER FOUR
===========
you never know when or if harry will at some future time use a quagmire four cipher, so having solved this puts you in good stead and readyness should the occasion arise.
i know that this text is not really long enough to do an analysis on to tell what type of cipher it might be that is why i gave you all the clues to it in cipher three – queue for (q4) and the three keywords were given. i did not intend you to have to find those but rather to have the ability to simply solve it, which you must have done to read this, so well done you!
if you want to post to prove that you have solved all the ciphers print the last word from each of the three preceding ciphers followed by a number being the count of the characters in your user name, for example ‘tlw mini word word word ninteen’ would be submitted by the-letter-wriggler.

ALL THE CIPHER TYPES AND KEYS
=============================
CIPHER ONE
==========
TYPE : Caesar
KEY : Shift of 13

CIPHER TWO
==========
TYPE : Mono-Alphabetic Substituiton
KEY : SOLVE
[PT] : abcdefghijklmnopqrstuvwxyz
[CT] : SOLVEABCDFGHIJKMNPQRTUWXYZ

CIPHER THREE
============
TYPE : Vigenere
CODE : Standard
KEYS : a-z, A-Z, REVEALING, a
[1] : abcdefghijklmnopqrstuvwxyz
[2] : ABCDEFGHIJKLMNOPQRSTUVWXYZ
[3] : REVEALING (Length of 9)
[4] : a (Start [1] with this letter)
[5] : Text was taken as a whole (carriage returns ignored)
IKey : *Auto Key Not Used*
—————————–
Key to above:
TYPE : The Cipher Type
CODE : Standard, Beaufort or Variant Beaufort
KEYS : Actual Keywords Used
[1] : Plaintext Alpha
[2] : 00 Matrix Alpha
[3] : Keyword
[4] : Plaintext Shift
[5] : The whole, or line by line, how text is taken
IKey : Intermediate Key (Auto PT/CT)
——————————-
Key Tableau Used(8 Shifts)
-> abcdefghijklmnopqrstuvwxyz
00 ABCDEFGHIJKLMNOPQRSTUVWXYZ
04 EFGHIJKLMNOPQRSTUVWXYZABCD
06 GHIJKLMNOPQRSTUVWXYZABCDEF
08 IJKLMNOPQRSTUVWXYZABCDEFGH
11 LMNOPQRSTUVWXYZABCDEFGHIJK
13 NOPQRSTUVWXYZABCDEFGHIJKLM
17 RSTUVWXYZABCDEFGHIJKLMNOPQ
21 VWXYZABCDEFGHIJKLMNOPQRSTU
——————————-

CIPHER FOUR
===========
TYPE : Quagmire IV
CODE : Standard
KEYS : cunning, INSIGHT, PERSEVERANCE, c
[1] : cunigabdefhjklmopqrstvwxyz
[2] : INSGHTABCDEFJKLMOPQRUVWXYZ
[3] : PERSEVERANCE (Length of 12)
[4] : c (Start [1] with this letter)
[5] : Text was taken as a whole (carriage returns ignored)
IKey : *Auto Key Not Used*
—————————–
Key to above:
TYPE : The Cipher Type
CODE : Standard, Beaufort or Variant Beaufort
KEYS : Actual Keywords Used
[1] : Plaintext Alpha
[2] : 00 Matrix Alpha
[3] : Keyword
[4] : Plaintext Shift
[5] : The whole, or line by line, how text is taken
IKey : Intermediate Key (Auto PT/CT)
——————————-
Key Tableau Used(8 Shifts)
-> cunigabdefhjklmopqrstvwxyz
01 NSGHTABCDEFJKLMOPQRUVWXYZI
02 SGHTABCDEFJKLMOPQRUVWXYZIN
06 ABCDEFJKLMOPQRUVWXYZINSGHT
08 CDEFJKLMOPQRUVWXYZINSGHTAB
10 EFJKLMOPQRUVWXYZINSGHTABCD
17 PQRUVWXYZINSGHTABCDEFJKLMO
19 RUVWXYZINSGHTABCDEFJKLMOPQ
21 VWXYZINSGHTABCDEFJKLMOPQRU
——————————-
The Full Tableau:
-> cunigabdefhjklmopqrstvwxyz
00 INSGHTABCDEFJKLMOPQRUVWXYZ
01 NSGHTABCDEFJKLMOPQRUVWXYZI
02 SGHTABCDEFJKLMOPQRUVWXYZIN
03 GHTABCDEFJKLMOPQRUVWXYZINS
04 HTABCDEFJKLMOPQRUVWXYZINSG
05 TABCDEFJKLMOPQRUVWXYZINSGH
06 ABCDEFJKLMOPQRUVWXYZINSGHT
07 BCDEFJKLMOPQRUVWXYZINSGHTA
08 CDEFJKLMOPQRUVWXYZINSGHTAB
09 DEFJKLMOPQRUVWXYZINSGHTABC
10 EFJKLMOPQRUVWXYZINSGHTABCD
11 FJKLMOPQRUVWXYZINSGHTABCDE
12 JKLMOPQRUVWXYZINSGHTABCDEF
13 KLMOPQRUVWXYZINSGHTABCDEFJ
14 LMOPQRUVWXYZINSGHTABCDEFJK
15 MOPQRUVWXYZINSGHTABCDEFJKL
16 OPQRUVWXYZINSGHTABCDEFJKLM
17 PQRUVWXYZINSGHTABCDEFJKLMO
18 QRUVWXYZINSGHTABCDEFJKLMOP
19 RUVWXYZINSGHTABCDEFJKLMOPQ
20 UVWXYZINSGHTABCDEFJKLMOPQR
21 VWXYZINSGHTABCDEFJKLMOPQRU
22 WXYZINSGHTABCDEFJKLMOPQRUV
23 XYZINSGHTABCDEFJKLMOPQRUVW
24 YZINSGHTABCDEFJKLMOPQRUVWX
25 ZINSGHTABCDEFJKLMOPQRUVWXY
——————————-

====================================================================
TLW’s A, B, C quizz. Can you answer these?
=========================================================== A ANSWER
GETTING ROPED IN
=================
The Length Of The Rope Was 5 Feet.

When the mother is three times as old as the monkey, let
their respective ages be 3x and x.
Thus the difference of their ages is 3x-1x = 2x(constant).
When the monkey is tree times as old as the mother then was, he will be 9x.
When the mother is half this age, she is 4.5x, and the monkey therefore is 2.5x.
Twice this i.e. 5x, is the mother’s age at the time of the problem, and the
monkey’s age is therefore 3x.
Now 5x + 3x is to equal 4, therefore x = 0.5, and 5x = 2.5,
therefore weight of monkey = weight of weight = 2.5 pounds = 40 oz.
And we are told (expressing all weights in ounces)
40 + weight of rope = 1.5(40 + 40 – 40) = 60,
therefore weight of rope = 60 – 40 = 20 oz,
Rope is 4oz per foot so 20/4 = 5
and therefore length of rope = 5 feet.

============================================================ B ANSWER
GOING BANANAS
==============
The Banana is Five and Three Quarter Inches Long.

No write up for this solution, sorry.

============================================================ C ANSWER
Give the missing number N in this sequence.
===========================================
N is 121

The decimal number 16 is given in each Base (16 down to 1)
So N is 121 being decimal 16 in Base 3

This integer sequence can be checked out at https://oeis.org

====================================================================
============================================ ANSWER
TLW A Six Fingered Alien.
==========================
Here is my answer. This is how I worked it out, there are other ways.

First calculte how many unique numbers are possible

12! 12!
———— = ———– = 74484400 unique numbers
2!2!2!2!2!2! 2^6

(so we have 12 digit columns having 74484400 rows of numbers)

Now there can only be six different digits (1 2 3 4 5 and 6) in each
of the 12 columns that recur so:

74484400/6 = 1247400 recurences of the 6 digits.

With 1+2+3+4+5+6 = 21 the units column will total to 124700*21 = 26195400

This means each of the 12 columns will total 26195400*10^columns_power.

We can do this using Repunit(12) = 111111111111

26195400*111111111111 = 2910599999997089400

So my full workings are: ((12!/2^6)/6)*21*111111111111 = 29105999999999997089400

In number form:
479001600/64/6 = 1247400, 1247400*21 = 26195400, 26195400*111111111111 = 2910599999997089400

SEE THE NEXT ENTRY BELOW FOR MORE ON THIS FROM A COMPETITOR (MADNESS)
====================================================================
I WOULD LIKE TO INCLUDE WITH THESE ANSWERS THE FOLLOWING POST BY
MADNESS IN RESPONSE TO TLW A Six Fingered Alien. I like his thinking.

28th April 2020 at 3:54 pm #47945
Madness
Member

@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.

====================================================================
============================================ ANSWER
TLW The 10 Digit ‘Tally Number’
================================

The answer is 6210001000
========================
Meaning:
0] 6 the digit 0 appears 6 times
1] 2 the digit 2 appears once
2] 1 the digit 1 appears twice
3] 0 the digit 3 appears 0 times
4] 0 the digit 4 appears 0 times
5] 0 the digit 5 appears 0 times
6] 1 the digit 6 appears once
7] 0 the digit 7 appears 0 times
8] 0 the digit 8 appears 0 times
9] 0 the digit 9 appears 0 times

====================================================================
============================================ ANSWER
TLW The 10 Digit ‘Tally Number’ Continued
============================================
Note: No ‘tally number’ exists for bases 1,2,3 and 6.

[0123] base 4 two solutions
[1210]
[2020]

[01234] base 5 one solution
[21200]

For all bases above 6 there is just one solution.

In general using concatination of the bracketed sets we can use the following

[n][21][0..0][1000] (Martin Gardner called this the self-descriptive number)

where n is 4 less than the base,
the number of zero digits to include [0..0], is 7 less than the base,
and the 21 and 1000 are fixed digits.

So if b is the base number (higher that 6) then we calcuate thus:
[b-4][21][number of zeros = b-7][1000]

Working this for base 10 gives:
[n=10-4=6][21][10-7= 3 zeros][1000] so we have 6 21 000 1000 = 6210001000

and for other bases

[012345] base 7 [7-4][21][7-7][1000]=[3][21][][1000]
[301000]

[01234567] base 8 [8-4][21][8-7][1000]=[4][21][1][1000]
[42101000]

[012345678] base 9 [9-4][21][9-7][1000]=[5][21][2][1000]
[521001000]

[0123456789a] base 11 [11-4][21][11-7][1000]=[7][21][4][1000]
[72100001000]

[0123456789ab] base 12 [12-4][21][12-7][1000]=[8][21][5][1000]
[821000001000]

[0123456789abc] base 13 [13-4][21][13-7][1000]=[9][21][6][1000]
[9210000001000]

Double digit numbers or greater can be put like so:

Base 36 [36-4][21][0’s: 36-7][1000] (32)21(29)1000

[012345678901234567890123456789012345]decimal count
[01233456789abcdefghijklmnopqrstuvwxy]
[v21000000000000000000000000000001000]

(note that v represents 32(in decimal) and that it appears once
and there are v zeros (32 zeros))

==================================================
The other two results of T.O. e Silva in full are:
Longest path before hitting a loop…
R1=0000122245
R2=2222044430 (there are 3 other solutions)
R3=1111300023 (there is 1 other solution)
R4=3333411104 (there is 1 other solution)
R5=0000122241 (there is 1 other solution)
R6=2222044440 (there are 2 other solutions)
R7=0000800008
R8=2222022220
R9=0000800008

and the longest length loop found…
R1=0001112444
R2=1111444033 (there is 1 other solution)
R3=2222000311 (there is 1 other solution)
R4=0000111244 (there is 1 other solution)
R5=0000111244 (there is 1 other solution)

================================================================
============================================ ANSWER
TLW The 27 Tiles Problem
=========================
The 3 Solutions

[1] 191618257269258476354938743

[2] 191218246279458634753968357

[3] 181915267285296475384639743

And the 3 reversals of course

[4] 347839453674852962752816191

[5] 753869357436854972642812191

[6] 347936483574692582762519181

The sums you would have posted:
[1]+[4] = 539457710944111439107754934
[2]+[5] = 945087603716313607396780548
[3]+[6] = 529851750859989058147158924

================================================================
============================================ ANSWER
TLW Alphametic Sum
===================
>CROSS
>ROADS+
======
DANGER

For base 10
There are 9 different letters
Giving One Unique Solution

S R D E O A G N C
3 6 1 4 2 5 7 8 9

>96233
>62513
======
158746

================================================================
============================================== ANSWER
TLW The Czech Logical Labyrinth. The solution:
==============================================

01,10,15,04,22,11,07,14,24,03,20,21,09,16,05,19,13,06,18,08,12,02,23,17,25.

[01]_02__07_[16][05][19]_07__08
[10]_17__23_[09][21][13][06]_23
[15][04][22]_17_[20]_04_[18]_22
_19__17_[11][24][03]_17_[08]_19
_12__09_[07][14]_23__18_[12]_06
_06__05__23__14__10__05_[02][23]
_08__20__10__11__19__18__01_[17]
_02__24__15__22__13__04__04_[25]

================================================================

@Madness, thanks for pointing out the fatal flaw! I should have given s=ut+(1/2)at^2!

[10degrees-admin]

Wait, what did I write? Let’s try again.

The correct equation relating the fall time (t) and fall distance (x) of an object released at rest is x=(1/2)gt^2

[10degrees-admin]

I did put ‘x=gt^2’! [Did you need a factor of 1/2? Harry]

[10degrees-admin]

TLW The Hutchinson Problem
I have your 6-move way, @The-letter-wriggler!
2D 3R 1R 3D 2L 3R

[10degrees-admin]

TLW THE TRIPLET TRIP

There are only two solutions:
17 18 19 20 01 02
16 23 22 21 04 03
15 24 11 10 05 06
14 13 12 09 08 07

08 07 06 05 24 23
09 02 03 04 21 22
10 01 14 15 20 19
11 12 13 16 17 18

TLW THE HUTCHINSON PROBLEM

The fewest number of jumps I can find is 6: 2D-3R-1R-3D-2L-3R

[10degrees-admin]

KA-7 – The Second Riddle of The Sphinx

The title says it all this time. Enjoy!

There are two sisters. One gives birth to the other one, who in turn gives birth to the first. Who are they?

[Author]

Posts