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Nice puzzle @Kford-academy – I get “KA-5 86319720”?
KA Snack Place #1 Correct
=========================
Your bottom right quadrant differed from (mine), that can happen.
If you post more solutions please supply the directions as here,
it is very difficult to follow each intended snake for checking.[4][3][X][4][1][2] R1 1>RDD
[1][2][1][3][X][3] R2 1>DRR,1>LUL
[2][3][4][2][1][4] R3 1>LUU
[4][1][2][1][2][X] R4 1>RDD,1>RDR(,1>DDR)
[3][X][3][4][3][4] R5
[2][1][4][3][2][1] R6 1>LUU,1>LLU(,1>ULU)KA-5 86319720
KA-5 Answer
Well done to everyone who got the answer of ‘86319720’. Here is a reminder of the clues (I will refer to these throughout my method – for those who didn’t get it):
1. E-D=A
2. EH+E=A+D
3. DG=H+G
4. CG-G=G(G+H)
5. C+G=A-C
6. A=5+C
7. E^2=10A+1
8. B-A=D-C=F-E=H-GThe first step was to work out which letter H was. From Clue 1, E could not be 0 (as A would have been a negative number otherwise), so it was safe to divide by E in Clue 2. Rearranging Clue 1 makes E=A+D. So E can be substituted for ‘A+D’ in Clue 2, which comes up as EH+E=E. Dividing by E gets us H+1=1, which means H is 0. Substitute H for 0 in Clue 3 and you get DG=G. Dividing by G (as G isn’t 0) gets us D=1, our next letter. Substitute H for 0 again and in Clue 4 and you get CG-G=G(G). Divide by G and you get C-1=G. Substitute G for ‘C-1’ in Clue 5 and you get C+C-1=A-C. Adding C to both sides and simplifying gets 3C-1=A, which in turn makes Clue 6 (after substituting A for ‘3C-1’) 3C-1=5+C. This can be rearranged to make 2C=6, or C=3. Putting this into Clue 6 gets us A=5+3, or A=8. Doing the same to Clue 5 gets 3+G=8-3, or 3+G=5. This means that G is 2. In Clue 7, we can replace A for 8, giving us E^2 = 10*8+1, or E^2=81. So E=9. The only letters remaining are now B and F. In Clue 8, we are given a statement connecting all 8 letters. D-C is 1-3, or -2. So we can say that B-A=-2 and F-E=-2. Let’s start with B-A=-2 first. A is 8, so B-8=-2, or B=6. Now for F-E=-2. E is 9, so F-9=-2, or F=7. Just to summarize the answers: A is 8, B is 6, C is 3, D is 1, E is 9, F is 7, G is 2 and H is 0.
By all means post your method if you used a different one to mine!
I used clue 7, as the only square which is one greater than a multiple of ten is 81. If we use this fact then I think several of the clues become redundant.
There is a lot of redundancy in KA-5. It is possible to solve it using only clues 5, 7 and 8 (but any other set of three clues has multiple solutions).
B and F are only mentioned in the last clue; if we ignore those two letters (and delete the corresponding parts of clue 8), then it turns out that every single clue is redundant, and the only pair you can remove to get multiple possible solutions is clues 4 and 5.
If we count clue 8 as three separate equations (in some combination) then we need at least 5 clues to get a unique solution.
(Testing involved heavy application of itertools and lambdas in Python.)
Take a look at this https://twitter.com/kyledevans/status/1270968743816986626
18 Hours! It took me 1 minute – er providing I have the right answer.
I get:
Amy was 18 when she birthed the triplets and 9 years later when she is 27 the triplets combined age is 3×9 =27
or am I missing something?@The-letter-wriggler, the correct answer is that there is a 1/2 chance. (You just gave an example, which is NOT what the question asled for.) If Amy was an odd age when she birthed the triplets, it wouldn’t work; but if Amy was an even age when she birthed the triplets, it would work. Proof below.
When Amy birthed the triplets, her age was ‘A’, let’s say. After a certain period of B years:
A+B=3B
Which can be rearranged as:
A=2B
B must be a whole number. Therefore, A must be an even number – and obviously, there is a 1/2 chance that A is an even number. Took me about 30 seconds to get this answer – and I only looked at the solution afterwards!
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Snake Place Puzzle Number 2 complete! (With directions so you can check it.)
[5][4][1][X][1][2][1] R1 1>DLUL, 1>RDLL, 1>DDDL
[5][3][2][5][4][3][2] R2
[4][X][1][5][5][4][3] R3 1>DDDD
[3][1][2][4][3][5][4] R4 1>DDDL
[2][2][3][1][2][X][3] R5 1>RULU
[1][3][4][1][2][1][2] R6 1>UUUU, 1>RDRR, 1>RUUL
[5][4][5][X][3][4][5] R7TLW High Score Self Made Crosswords. My sixth grid – and better than before! Watch out, @The-letter-wriggler – I’m coming to get you!
[Z][Y][Z][Z][Y][V][A]
[Y][_][I][_][A][_][M]
[Z][I][Z][A][N][I][A]
[Z][_][A][_][K][_][L]
[Y][A][N][K][I][N][G]
[V][_][I][_][I][_][A]
[A][M][A][L][G][A][M]Score: 140 (25 points off the highest score, set by @The-letter-wriggler)
@Kford-academy
Yes, good one isn’t it. I never looked to see if it was already answered.
I gave that answer for a laugh! Knew it would draw a post.Your Snake Place #2 is right. Good aren’t they.
Very funny. Here is my response in the same cipher:
HHHCHIBHKIBEBGHCBBGFKEAEBAGJCJEKDJICKKABFEEKCCDFKDCDBEIEJGJEBDKIBF
GIKADGBEAAJADDCKEBFHCFFGIJDKKHKGFGDGKJJDGGEIEJCDBKFGEGJGKKIEECAEDF
GAGTLW A NUMBER RING
=================
Having just now encountered this…THE RING
[_][3][_][2]
[3][X][X][_]
[_][X][X][1]
[6][_][1][_]Going clockwise or counterclockwise the ring of numbers
form a mathematical expression when certain empty cells
are filled with + – / * ^ = and an E for End.
If you leave a blank between digits then the digits
concatenate to form a multi-digit number.
All digits have to be used.…here are six solutions I did in 15 minutes.
33*2/11=6
[_][3][*][2]
[3][X][X][/]
[E][X][X][1]
[6][=][1][_]1*6*3+3=21
[+][3][=][2]
[3][X][X][_]
[*][X][X][1]
[6][*][1][E]3+3-2+1+1=6
[+][3][-][2]
[3][X][X][+]
[E][X][X][1]
[6][=][1][+]3^3-21*1=6
[^][3][-][2]
[3][X][X][_]
[E][X][X][1]
[6][=][1][*]3*3-6-1*1=2
[*][3][E][2]
[3][X][X][=]
[-][X][X][1]
[6][-][1][*]6*11/2=33
[_][3][=][2]
[3][X][X][/]
[E][X][X][1]
[6][*][1][_]Try making as many more as you can if you wish.
These may or may not be trivial to come up with but…What I would like to know (I do not have an answer) is the limit, what
is the maximum number of such sums that can be made from this ring?
Would a ring, retaining this format, with different digits produce more or less sums?Sorry for taking so long to respond but have you used a different cipher in addition to the original to give your answer? I tried decrypting using my script and it just gave me a random string of characters. Whilst I am sure that you have successfully decoded the message, it would be nice to know what your message says. Sorry if I am missing something obvious.
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