A cipher for people who don't normally enjoy ciphers
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I've never been a fan of ciphers/encryption. Truth be told, despite my love of puzzles (and the amount of time I spend here on Puzzling.SE), I'm just no good at them and haven't properly learnt the strategies for trying to crack them.
So in an effort to expand on the types of ciphers and hopefully broaden the audience, I offer you the following challenge:
I have encrypted a five-word phrase in the form
_ _ _ _/ _ _ _ _, / _ _ _/ _ _ _ _ _ _/ _ _ _ _ _ _
[word lengths are (4) (4), (3) (6) (6)]
However, instead of just encrypting it once, I have done so six times, using six different methods.
The output for each encryption is as follows:
The final answer has two parts to it. First, the completed five-word phrase (which I suspect will be discovered first) and second, the six different encryption methods, all of which must be detailed in the answer.
The decrypted phrase will tell you to do something, so make sure you do it :) You wouldn't want to make the puzzle sad, would you?
Some excellent community effort so far! The phrase and methods 2, 4 and 5 have been cracked, so I'll throw some subtle hints for the remaining three methods in
Method 3
How often do you see a 3-letter word have a higher value than a 4-letter word?
Method 6
The order of the letters doesn't seem logical in isolation, but if it ever got changed, many of us would struggle to adapt!
cipher
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
$endgroup$
|
show 2 more comments
$begingroup$
I've never been a fan of ciphers/encryption. Truth be told, despite my love of puzzles (and the amount of time I spend here on Puzzling.SE), I'm just no good at them and haven't properly learnt the strategies for trying to crack them.
So in an effort to expand on the types of ciphers and hopefully broaden the audience, I offer you the following challenge:
I have encrypted a five-word phrase in the form
_ _ _ _/ _ _ _ _, / _ _ _/ _ _ _ _ _ _/ _ _ _ _ _ _
[word lengths are (4) (4), (3) (6) (6)]
However, instead of just encrypting it once, I have done so six times, using six different methods.
The output for each encryption is as follows:
The final answer has two parts to it. First, the completed five-word phrase (which I suspect will be discovered first) and second, the six different encryption methods, all of which must be detailed in the answer.
The decrypted phrase will tell you to do something, so make sure you do it :) You wouldn't want to make the puzzle sad, would you?
Some excellent community effort so far! The phrase and methods 2, 4 and 5 have been cracked, so I'll throw some subtle hints for the remaining three methods in
Method 3
How often do you see a 3-letter word have a higher value than a 4-letter word?
Method 6
The order of the letters doesn't seem logical in isolation, but if it ever got changed, many of us would struggle to adapt!
cipher
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
$endgroup$
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Shouldn't the phrase have 4 words?
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– Display name
Nov 12 '18 at 23:09
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Oops! - thanks for catching that!
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– Dmihawk
Nov 12 '18 at 23:28
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I have an idea for Method 4, but only if word 1 is 35 instead of 45. thanks!
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– Omega Krypton
Nov 13 '18 at 10:12
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I'll need to double check in the morning
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– Dmihawk
Nov 13 '18 at 10:18
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Sorry I was wrong
$endgroup$
– Omega Krypton
Nov 13 '18 at 23:34
|
show 2 more comments
$begingroup$
I've never been a fan of ciphers/encryption. Truth be told, despite my love of puzzles (and the amount of time I spend here on Puzzling.SE), I'm just no good at them and haven't properly learnt the strategies for trying to crack them.
So in an effort to expand on the types of ciphers and hopefully broaden the audience, I offer you the following challenge:
I have encrypted a five-word phrase in the form
_ _ _ _/ _ _ _ _, / _ _ _/ _ _ _ _ _ _/ _ _ _ _ _ _
[word lengths are (4) (4), (3) (6) (6)]
However, instead of just encrypting it once, I have done so six times, using six different methods.
The output for each encryption is as follows:
The final answer has two parts to it. First, the completed five-word phrase (which I suspect will be discovered first) and second, the six different encryption methods, all of which must be detailed in the answer.
The decrypted phrase will tell you to do something, so make sure you do it :) You wouldn't want to make the puzzle sad, would you?
Some excellent community effort so far! The phrase and methods 2, 4 and 5 have been cracked, so I'll throw some subtle hints for the remaining three methods in
Method 3
How often do you see a 3-letter word have a higher value than a 4-letter word?
Method 6
The order of the letters doesn't seem logical in isolation, but if it ever got changed, many of us would struggle to adapt!
cipher
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
$endgroup$
I've never been a fan of ciphers/encryption. Truth be told, despite my love of puzzles (and the amount of time I spend here on Puzzling.SE), I'm just no good at them and haven't properly learnt the strategies for trying to crack them.
So in an effort to expand on the types of ciphers and hopefully broaden the audience, I offer you the following challenge:
I have encrypted a five-word phrase in the form
_ _ _ _/ _ _ _ _, / _ _ _/ _ _ _ _ _ _/ _ _ _ _ _ _
[word lengths are (4) (4), (3) (6) (6)]
However, instead of just encrypting it once, I have done so six times, using six different methods.
The output for each encryption is as follows:
The final answer has two parts to it. First, the completed five-word phrase (which I suspect will be discovered first) and second, the six different encryption methods, all of which must be detailed in the answer.
The decrypted phrase will tell you to do something, so make sure you do it :) You wouldn't want to make the puzzle sad, would you?
Some excellent community effort so far! The phrase and methods 2, 4 and 5 have been cracked, so I'll throw some subtle hints for the remaining three methods in
Method 3
How often do you see a 3-letter word have a higher value than a 4-letter word?
Method 6
The order of the letters doesn't seem logical in isolation, but if it ever got changed, many of us would struggle to adapt!
cipher
cipher
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
edited Nov 14 '18 at 18:51
Dmihawk
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
asked Nov 12 '18 at 22:29
DmihawkDmihawk
1,966523
1,966523
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Shouldn't the phrase have 4 words?
$endgroup$
– Display name
Nov 12 '18 at 23:09
$begingroup$
Oops! - thanks for catching that!
$endgroup$
– Dmihawk
Nov 12 '18 at 23:28
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I have an idea for Method 4, but only if word 1 is 35 instead of 45. thanks!
$endgroup$
– Omega Krypton
Nov 13 '18 at 10:12
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I'll need to double check in the morning
$endgroup$
– Dmihawk
Nov 13 '18 at 10:18
$begingroup$
Sorry I was wrong
$endgroup$
– Omega Krypton
Nov 13 '18 at 23:34
|
show 2 more comments
$begingroup$
Shouldn't the phrase have 4 words?
$endgroup$
– Display name
Nov 12 '18 at 23:09
$begingroup$
Oops! - thanks for catching that!
$endgroup$
– Dmihawk
Nov 12 '18 at 23:28
$begingroup$
I have an idea for Method 4, but only if word 1 is 35 instead of 45. thanks!
$endgroup$
– Omega Krypton
Nov 13 '18 at 10:12
$begingroup$
I'll need to double check in the morning
$endgroup$
– Dmihawk
Nov 13 '18 at 10:18
$begingroup$
Sorry I was wrong
$endgroup$
– Omega Krypton
Nov 13 '18 at 23:34
$begingroup$
Shouldn't the phrase have 4 words?
$endgroup$
– Display name
Nov 12 '18 at 23:09
$begingroup$
Shouldn't the phrase have 4 words?
$endgroup$
– Display name
Nov 12 '18 at 23:09
$begingroup$
Oops! - thanks for catching that!
$endgroup$
– Dmihawk
Nov 12 '18 at 23:28
$begingroup$
Oops! - thanks for catching that!
$endgroup$
– Dmihawk
Nov 12 '18 at 23:28
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I have an idea for Method 4, but only if word 1 is 35 instead of 45. thanks!
$endgroup$
– Omega Krypton
Nov 13 '18 at 10:12
$begingroup$
I have an idea for Method 4, but only if word 1 is 35 instead of 45. thanks!
$endgroup$
– Omega Krypton
Nov 13 '18 at 10:12
$begingroup$
I'll need to double check in the morning
$endgroup$
– Dmihawk
Nov 13 '18 at 10:18
$begingroup$
I'll need to double check in the morning
$endgroup$
– Dmihawk
Nov 13 '18 at 10:18
$begingroup$
Sorry I was wrong
$endgroup$
– Omega Krypton
Nov 13 '18 at 23:34
$begingroup$
Sorry I was wrong
$endgroup$
– Omega Krypton
Nov 13 '18 at 23:34
|
show 2 more comments
4 Answers
4
active
oldest
votes
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This is a summary of all answers provided by everyone, I have credited them, if I left anyone out, please state it in the comments. Thanks!
Partial Answer:
The phrase is:
WELL DONE, NOW UPVOTE PUZZLE (which I did), thanks to @DrXorile (approved by OP in comment)
Method 1 is
Sum of Scrabble values of each letter. So much thanks to @Braegh! (approved by OP in comment)
My original guess:
Ceasar shifting, the number being the rotation number, as all numbers are not larger than 26. 26 means the word is not shifted at all. (I believe that this is wrong...)
Method 2 is:
Sum of values in A1Z26 scheme, Thanks to @ImongMama (approved by OP in comment)
Method 3 is:
The sum of values of each letter according to the frequency (descending): etaoi nshrd lcumw fgypb vkjxq z, where e=1, z=26,
"WELL"=15+1+11+11=38
THANKS TO @NudgeNudge!!
Method 4 is:
the sum of values of each letter on a telephone keypad, like this:
e.g. for "DONE" 3+(6+6+6)+(6+6)+(3+3)=39 (approved by OP in comment)
Method 5 is
the sum of ASCII values of each letter, since they add up to around 80-90 per letter. (approved by OP in comment)
And method 6:
is the sum of the values for each letter according to its position on a QWERTY keyboard, like this:
PS:
I really enjoyed this puzzle. Thanks so much, @Dmihawk!
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
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Good spotting! :)
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– Dmihawk
Nov 12 '18 at 23:41
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Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other...
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– Dmihawk
Nov 13 '18 at 9:42
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Method 6? @Dmihawk Thanks!
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– Omega Krypton
Nov 13 '18 at 9:57
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Sorry, meant 5!
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– Dmihawk
Nov 13 '18 at 10:14
2
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For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...))
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– NudgeNudge
Nov 14 '18 at 20:57
|
show 3 more comments
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Method 2 is
Sum of A1Z26 values
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
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add a comment |
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I believe Method 1 is
the (English) Scrabble value for each word.
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
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Well done! Good find :)
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– Dmihawk
Nov 14 '18 at 0:35
add a comment |
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Based on @Omega Krypton's method 5, the phrase is:
WELL DONE, NOW UPVOTE PUZZLE
I figured this out by going through all the english words that would fit those sums. The last word stood out, and I got the remaining ones except for the fourth quite quickly. The fourth wasn't in my dictionary, but easy enough to guess and check...
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
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Nicely done :) now you just need to deduce the other 5 encryption methods!
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– Dmihawk
Nov 13 '18 at 0:37
1
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I upvoted the puzzle lah :)
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– Omega Krypton
Nov 13 '18 at 2:57
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a summary of all answers provided by everyone, I have credited them, if I left anyone out, please state it in the comments. Thanks!
Partial Answer:
The phrase is:
WELL DONE, NOW UPVOTE PUZZLE (which I did), thanks to @DrXorile (approved by OP in comment)
Method 1 is
Sum of Scrabble values of each letter. So much thanks to @Braegh! (approved by OP in comment)
My original guess:
Ceasar shifting, the number being the rotation number, as all numbers are not larger than 26. 26 means the word is not shifted at all. (I believe that this is wrong...)
Method 2 is:
Sum of values in A1Z26 scheme, Thanks to @ImongMama (approved by OP in comment)
Method 3 is:
The sum of values of each letter according to the frequency (descending): etaoi nshrd lcumw fgypb vkjxq z, where e=1, z=26,
"WELL"=15+1+11+11=38
THANKS TO @NudgeNudge!!
Method 4 is:
the sum of values of each letter on a telephone keypad, like this:
e.g. for "DONE" 3+(6+6+6)+(6+6)+(3+3)=39 (approved by OP in comment)
Method 5 is
the sum of ASCII values of each letter, since they add up to around 80-90 per letter. (approved by OP in comment)
And method 6:
is the sum of the values for each letter according to its position on a QWERTY keyboard, like this:
PS:
I really enjoyed this puzzle. Thanks so much, @Dmihawk!
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
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$begingroup$
Good spotting! :)
$endgroup$
– Dmihawk
Nov 12 '18 at 23:41
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Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other...
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– Dmihawk
Nov 13 '18 at 9:42
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Method 6? @Dmihawk Thanks!
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– Omega Krypton
Nov 13 '18 at 9:57
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Sorry, meant 5!
$endgroup$
– Dmihawk
Nov 13 '18 at 10:14
2
$begingroup$
For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...))
$endgroup$
– NudgeNudge
Nov 14 '18 at 20:57
|
show 3 more comments
$begingroup$
This is a summary of all answers provided by everyone, I have credited them, if I left anyone out, please state it in the comments. Thanks!
Partial Answer:
The phrase is:
WELL DONE, NOW UPVOTE PUZZLE (which I did), thanks to @DrXorile (approved by OP in comment)
Method 1 is
Sum of Scrabble values of each letter. So much thanks to @Braegh! (approved by OP in comment)
My original guess:
Ceasar shifting, the number being the rotation number, as all numbers are not larger than 26. 26 means the word is not shifted at all. (I believe that this is wrong...)
Method 2 is:
Sum of values in A1Z26 scheme, Thanks to @ImongMama (approved by OP in comment)
Method 3 is:
The sum of values of each letter according to the frequency (descending): etaoi nshrd lcumw fgypb vkjxq z, where e=1, z=26,
"WELL"=15+1+11+11=38
THANKS TO @NudgeNudge!!
Method 4 is:
the sum of values of each letter on a telephone keypad, like this:
e.g. for "DONE" 3+(6+6+6)+(6+6)+(3+3)=39 (approved by OP in comment)
Method 5 is
the sum of ASCII values of each letter, since they add up to around 80-90 per letter. (approved by OP in comment)
And method 6:
is the sum of the values for each letter according to its position on a QWERTY keyboard, like this:
PS:
I really enjoyed this puzzle. Thanks so much, @Dmihawk!
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
$endgroup$
$begingroup$
Good spotting! :)
$endgroup$
– Dmihawk
Nov 12 '18 at 23:41
$begingroup$
Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other...
$endgroup$
– Dmihawk
Nov 13 '18 at 9:42
$begingroup$
Method 6? @Dmihawk Thanks!
$endgroup$
– Omega Krypton
Nov 13 '18 at 9:57
$begingroup$
Sorry, meant 5!
$endgroup$
– Dmihawk
Nov 13 '18 at 10:14
2
$begingroup$
For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...))
$endgroup$
– NudgeNudge
Nov 14 '18 at 20:57
|
show 3 more comments
$begingroup$
This is a summary of all answers provided by everyone, I have credited them, if I left anyone out, please state it in the comments. Thanks!
Partial Answer:
The phrase is:
WELL DONE, NOW UPVOTE PUZZLE (which I did), thanks to @DrXorile (approved by OP in comment)
Method 1 is
Sum of Scrabble values of each letter. So much thanks to @Braegh! (approved by OP in comment)
My original guess:
Ceasar shifting, the number being the rotation number, as all numbers are not larger than 26. 26 means the word is not shifted at all. (I believe that this is wrong...)
Method 2 is:
Sum of values in A1Z26 scheme, Thanks to @ImongMama (approved by OP in comment)
Method 3 is:
The sum of values of each letter according to the frequency (descending): etaoi nshrd lcumw fgypb vkjxq z, where e=1, z=26,
"WELL"=15+1+11+11=38
THANKS TO @NudgeNudge!!
Method 4 is:
the sum of values of each letter on a telephone keypad, like this:
e.g. for "DONE" 3+(6+6+6)+(6+6)+(3+3)=39 (approved by OP in comment)
Method 5 is
the sum of ASCII values of each letter, since they add up to around 80-90 per letter. (approved by OP in comment)
And method 6:
is the sum of the values for each letter according to its position on a QWERTY keyboard, like this:
PS:
I really enjoyed this puzzle. Thanks so much, @Dmihawk!
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
$endgroup$
This is a summary of all answers provided by everyone, I have credited them, if I left anyone out, please state it in the comments. Thanks!
Partial Answer:
The phrase is:
WELL DONE, NOW UPVOTE PUZZLE (which I did), thanks to @DrXorile (approved by OP in comment)
Method 1 is
Sum of Scrabble values of each letter. So much thanks to @Braegh! (approved by OP in comment)
My original guess:
Ceasar shifting, the number being the rotation number, as all numbers are not larger than 26. 26 means the word is not shifted at all. (I believe that this is wrong...)
Method 2 is:
Sum of values in A1Z26 scheme, Thanks to @ImongMama (approved by OP in comment)
Method 3 is:
The sum of values of each letter according to the frequency (descending): etaoi nshrd lcumw fgypb vkjxq z, where e=1, z=26,
"WELL"=15+1+11+11=38
THANKS TO @NudgeNudge!!
Method 4 is:
the sum of values of each letter on a telephone keypad, like this:
e.g. for "DONE" 3+(6+6+6)+(6+6)+(3+3)=39 (approved by OP in comment)
Method 5 is
the sum of ASCII values of each letter, since they add up to around 80-90 per letter. (approved by OP in comment)
And method 6:
is the sum of the values for each letter according to its position on a QWERTY keyboard, like this:
PS:
I really enjoyed this puzzle. Thanks so much, @Dmihawk!
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
edited Nov 14 '18 at 23:55
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
answered Nov 12 '18 at 23:40
Omega KryptonOmega Krypton
2,5301227
2,5301227
$begingroup$
Good spotting! :)
$endgroup$
– Dmihawk
Nov 12 '18 at 23:41
$begingroup$
Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other...
$endgroup$
– Dmihawk
Nov 13 '18 at 9:42
$begingroup$
Method 6? @Dmihawk Thanks!
$endgroup$
– Omega Krypton
Nov 13 '18 at 9:57
$begingroup$
Sorry, meant 5!
$endgroup$
– Dmihawk
Nov 13 '18 at 10:14
2
$begingroup$
For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...))
$endgroup$
– NudgeNudge
Nov 14 '18 at 20:57
|
show 3 more comments
$begingroup$
Good spotting! :)
$endgroup$
– Dmihawk
Nov 12 '18 at 23:41
$begingroup$
Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other...
$endgroup$
– Dmihawk
Nov 13 '18 at 9:42
$begingroup$
Method 6? @Dmihawk Thanks!
$endgroup$
– Omega Krypton
Nov 13 '18 at 9:57
$begingroup$
Sorry, meant 5!
$endgroup$
– Dmihawk
Nov 13 '18 at 10:14
2
$begingroup$
For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...))
$endgroup$
– NudgeNudge
Nov 14 '18 at 20:57
$begingroup$
Good spotting! :)
$endgroup$
– Dmihawk
Nov 12 '18 at 23:41
$begingroup$
Good spotting! :)
$endgroup$
– Dmihawk
Nov 12 '18 at 23:41
$begingroup$
Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other...
$endgroup$
– Dmihawk
Nov 13 '18 at 9:42
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Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other...
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– Dmihawk
Nov 13 '18 at 9:42
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Method 6? @Dmihawk Thanks!
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– Omega Krypton
Nov 13 '18 at 9:57
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Method 6? @Dmihawk Thanks!
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– Omega Krypton
Nov 13 '18 at 9:57
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Sorry, meant 5!
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– Dmihawk
Nov 13 '18 at 10:14
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Sorry, meant 5!
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– Dmihawk
Nov 13 '18 at 10:14
2
2
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For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...))
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– NudgeNudge
Nov 14 '18 at 20:57
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For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...))
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– NudgeNudge
Nov 14 '18 at 20:57
|
show 3 more comments
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Method 2 is
Sum of A1Z26 values
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
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add a comment |
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Method 2 is
Sum of A1Z26 values
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
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add a comment |
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Method 2 is
Sum of A1Z26 values
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
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Method 2 is
Sum of A1Z26 values
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
answered Nov 13 '18 at 7:06
ImongMamaImongMama
60519
60519
add a comment |
add a comment |
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I believe Method 1 is
the (English) Scrabble value for each word.
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
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Well done! Good find :)
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– Dmihawk
Nov 14 '18 at 0:35
add a comment |
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I believe Method 1 is
the (English) Scrabble value for each word.
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
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Well done! Good find :)
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– Dmihawk
Nov 14 '18 at 0:35
add a comment |
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I believe Method 1 is
the (English) Scrabble value for each word.
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
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I believe Method 1 is
the (English) Scrabble value for each word.
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
answered Nov 14 '18 at 0:34
BraeghBraegh
3537
3537
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Well done! Good find :)
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– Dmihawk
Nov 14 '18 at 0:35
add a comment |
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Well done! Good find :)
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– Dmihawk
Nov 14 '18 at 0:35
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Well done! Good find :)
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– Dmihawk
Nov 14 '18 at 0:35
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Well done! Good find :)
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– Dmihawk
Nov 14 '18 at 0:35
add a comment |
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Based on @Omega Krypton's method 5, the phrase is:
WELL DONE, NOW UPVOTE PUZZLE
I figured this out by going through all the english words that would fit those sums. The last word stood out, and I got the remaining ones except for the fourth quite quickly. The fourth wasn't in my dictionary, but easy enough to guess and check...
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
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Nicely done :) now you just need to deduce the other 5 encryption methods!
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– Dmihawk
Nov 13 '18 at 0:37
1
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I upvoted the puzzle lah :)
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– Omega Krypton
Nov 13 '18 at 2:57
add a comment |
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Based on @Omega Krypton's method 5, the phrase is:
WELL DONE, NOW UPVOTE PUZZLE
I figured this out by going through all the english words that would fit those sums. The last word stood out, and I got the remaining ones except for the fourth quite quickly. The fourth wasn't in my dictionary, but easy enough to guess and check...
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
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Nicely done :) now you just need to deduce the other 5 encryption methods!
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– Dmihawk
Nov 13 '18 at 0:37
1
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I upvoted the puzzle lah :)
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– Omega Krypton
Nov 13 '18 at 2:57
add a comment |
$begingroup$
Based on @Omega Krypton's method 5, the phrase is:
WELL DONE, NOW UPVOTE PUZZLE
I figured this out by going through all the english words that would fit those sums. The last word stood out, and I got the remaining ones except for the fourth quite quickly. The fourth wasn't in my dictionary, but easy enough to guess and check...
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
$endgroup$
Based on @Omega Krypton's method 5, the phrase is:
WELL DONE, NOW UPVOTE PUZZLE
I figured this out by going through all the english words that would fit those sums. The last word stood out, and I got the remaining ones except for the fourth quite quickly. The fourth wasn't in my dictionary, but easy enough to guess and check...
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
answered Nov 13 '18 at 0:36
Dr XorileDr Xorile
11.7k22566
11.7k22566
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Nicely done :) now you just need to deduce the other 5 encryption methods!
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– Dmihawk
Nov 13 '18 at 0:37
1
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I upvoted the puzzle lah :)
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– Omega Krypton
Nov 13 '18 at 2:57
add a comment |
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Nicely done :) now you just need to deduce the other 5 encryption methods!
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– Dmihawk
Nov 13 '18 at 0:37
1
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I upvoted the puzzle lah :)
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– Omega Krypton
Nov 13 '18 at 2:57
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Nicely done :) now you just need to deduce the other 5 encryption methods!
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– Dmihawk
Nov 13 '18 at 0:37
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Nicely done :) now you just need to deduce the other 5 encryption methods!
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– Dmihawk
Nov 13 '18 at 0:37
1
1
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I upvoted the puzzle lah :)
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– Omega Krypton
Nov 13 '18 at 2:57
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I upvoted the puzzle lah :)
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– Omega Krypton
Nov 13 '18 at 2:57
add a comment |
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Shouldn't the phrase have 4 words?
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– Display name
Nov 12 '18 at 23:09
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Oops! - thanks for catching that!
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– Dmihawk
Nov 12 '18 at 23:28
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I have an idea for Method 4, but only if word 1 is 35 instead of 45. thanks!
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– Omega Krypton
Nov 13 '18 at 10:12
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I'll need to double check in the morning
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– Dmihawk
Nov 13 '18 at 10:18
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Sorry I was wrong
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– Omega Krypton
Nov 13 '18 at 23:34