Sum letters are not two different Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Product and SumA mysterious property of 17 US-statesA fellowship of nine wordsAnother Company of ThirteenDetermine $X$ and $Y$: sum, difference and ratioWhen logic problems require “yes-no” questions, are there conventions for questions that might get an “I don't know”?Three mathematicians are forever in PrisonAn Only Connect WallWhat is the next integer in this sequence?Why is this sequence of images correct?
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Sum letters are not two different
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Product and SumA mysterious property of 17 US-statesA fellowship of nine wordsAnother Company of ThirteenDetermine $X$ and $Y$: sum, difference and ratioWhen logic problems require “yes-no” questions, are there conventions for questions that might get an “I don't know”?Three mathematicians are forever in PrisonAn Only Connect WallWhat is the next integer in this sequence?Why is this sequence of images correct?
$begingroup$
The following letters all have something in common which may not be obvious at a first glance:
A B D H P
No other letters share this attribute.
Hint
There are no misspellings or typos in the title of this question. Maybe a clue though.
More hints may follow if the question is not answered.
Some great answers so far all of which I have upvoted but none of which are exactly what I am looking for.
Hint 2
The number
0
and the character(
also have the same property. I only said no other letters share it ;-)
Hint 3
As correctly identified by @RedBaron the ASCII table is key here. There is a good reason why "sum" is mentioned in the title and there are two reasons why "two" is mentioned.
logical-deduction pattern
$endgroup$
add a comment |
$begingroup$
The following letters all have something in common which may not be obvious at a first glance:
A B D H P
No other letters share this attribute.
Hint
There are no misspellings or typos in the title of this question. Maybe a clue though.
More hints may follow if the question is not answered.
Some great answers so far all of which I have upvoted but none of which are exactly what I am looking for.
Hint 2
The number
0
and the character(
also have the same property. I only said no other letters share it ;-)
Hint 3
As correctly identified by @RedBaron the ASCII table is key here. There is a good reason why "sum" is mentioned in the title and there are two reasons why "two" is mentioned.
logical-deduction pattern
$endgroup$
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
2 days ago
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
Does a!
also share this property?
$endgroup$
– Eagle
2 days ago
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
2 days ago
add a comment |
$begingroup$
The following letters all have something in common which may not be obvious at a first glance:
A B D H P
No other letters share this attribute.
Hint
There are no misspellings or typos in the title of this question. Maybe a clue though.
More hints may follow if the question is not answered.
Some great answers so far all of which I have upvoted but none of which are exactly what I am looking for.
Hint 2
The number
0
and the character(
also have the same property. I only said no other letters share it ;-)
Hint 3
As correctly identified by @RedBaron the ASCII table is key here. There is a good reason why "sum" is mentioned in the title and there are two reasons why "two" is mentioned.
logical-deduction pattern
$endgroup$
The following letters all have something in common which may not be obvious at a first glance:
A B D H P
No other letters share this attribute.
Hint
There are no misspellings or typos in the title of this question. Maybe a clue though.
More hints may follow if the question is not answered.
Some great answers so far all of which I have upvoted but none of which are exactly what I am looking for.
Hint 2
The number
0
and the character(
also have the same property. I only said no other letters share it ;-)
Hint 3
As correctly identified by @RedBaron the ASCII table is key here. There is a good reason why "sum" is mentioned in the title and there are two reasons why "two" is mentioned.
logical-deduction pattern
logical-deduction pattern
edited 2 days ago
ElPedro
asked 2 days ago
ElPedroElPedro
303210
303210
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
2 days ago
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
Does a!
also share this property?
$endgroup$
– Eagle
2 days ago
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
2 days ago
add a comment |
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
2 days ago
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
Does a!
also share this property?
$endgroup$
– Eagle
2 days ago
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
2 days ago
5
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
2 days ago
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
2 days ago
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
2 days ago
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
2 days ago
1
1
$begingroup$
Does a
!
also share this property?$endgroup$
– Eagle
2 days ago
$begingroup$
Does a
!
also share this property?$endgroup$
– Eagle
2 days ago
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
2 days ago
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
2 days ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The property seems to be related to:
Binary equivalents of the symbols/alphabets etc.
Explanation:
Binary equivalents for the following can be written as:
A -- 01000001
B -- 01000010
D -- 01000100
H -- 01001000
P -- 01010000
0 -- 00110000
( -- 00101000
So the property is,
The sum of digits in the binary equivalents is two
Or
The binary equivalents of all these have two
1
s and six0
s.
Looking at the binary equivalents of the alphabets, one can see that no other alphabets share this property
The title (Thanks to @trolley813):
Two might refer to the sum of the digits, which is indeed two!
Title might mean that the sum [of digits in] letters is not different from two [but is equal to two]
Old (and wrong) answer
The property is:
The index if each alphabet is equal to the sum of indices of the preceding alphabets in the sequence +1.
And,
There is no other alphabet with the index
1+2+4+8+16+1 = 32
$endgroup$
1
$begingroup$
You've got it. Binary was what I was looking for but there are some other interesting patterns came out of this puzzle. I have added another hint in case anyone wants to have a go without looking at your answer. For the same reason, I'll wait a couple of hours before I accept it. well done!
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
@Eagle Some refinement about the title, it probably should read "sum [of digits in] letters is not different from two [i.e. equals to 2]"
$endgroup$
– trolley813
yesterday
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
yesterday
$begingroup$
@trolley813 - Close but actually more of a play on words. Rot13(Fhz (Fbzr) yrggref ner abg gjb (gbb) qvssrerag) with Rot13(Fhz) and Rot13(gjb) giving clues to what I was looking for ;-)
$endgroup$
– ElPedro
yesterday
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
$endgroup$
2
$begingroup$
That wasn't what I was looking for but is indeed true and is possibly a side effect of the answer that I was looking for so +1 but I won't mark it as accepted yet.
$endgroup$
– ElPedro
2 days ago
2
$begingroup$
@ElPedro: It is a side effect, but it hinges on rot13(jurer gur nycunorg fgnegf ba gur NFPVV gnoyr. Vtaber gur svefg ovg bs gur NFPVV inyhr (orpnhfr vg vf nyjnlf 1 naq arire punatrf sebz N gb M), ohg QB erzrzore gung vg nyjnlf pbagevohgrf gb bar bs gur gjb 1-ovgf va lbhe vagraqrq nafjre. Vtabevat gur svefg ovg, N rssrpgviryl fgnegf ng ahzrevpny inyhr 1, naq rirel "cbjre bs gjb" yrggre nqqf rknpgyl 1 1-ovg gb gur pbhag, juvpu rkcynvaf jul obgu nafjref ner pbeerpg. Vs N unq fgnegrq ba n qvssrerag NFPVV inyhr, vg znl abg unir orra gur pnfr.)
$endgroup$
– Flater
yesterday
$begingroup$
@Flater - Thanks for the explanation. It makes complete sense. As I said, some pretty interesting things have come out of what I at first though was a pretty simple puzzle :)
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
Is it
All the letters, symbols in this group have ASCII codes of form $2^m + 2^n$ where m and n are integers
Thus we have
From ascii code table,
$A = 65 = 64 + 1 = 2^6 + 2^0$
$B = 66 = 64 + 2 = 2^6 + 2^1$
$D = 68 = 64 + 4 = 2^6 + 2^2$
$H = 72 = 64 + 8 = 2^6 + 2^3$
$P = 80 = 64 + 16 = 2^6 + 2^4$
Other letters don't share this property because
64 + 32 = 96 which does not correspond to any letter. The letter
a
begins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
$endgroup$
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
2 days ago
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
2 days ago
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
I like the format of this answer more than the accepted one, simply because that's how I internally rephrased the accepted answer before even reading this one :) +1
$endgroup$
– Flater
yesterday
add a comment |
$begingroup$
I think it's:
Each letter's alphanumeric value is double its predecessor, which also means, sum two times the alphanumeric value of the previous letter
This means that:
Starting from A=1 we get the sequence 1,2,4,8,16,... which corresponds to the sequence A,B,D,H,P
New contributor
$endgroup$
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
2 days ago
add a comment |
$begingroup$
Each time you add the position of the letter (A is 1 and B is 2), the next letter's position is the sum of the previous +1.
Thus, 1+2 is 3, +4 is 7, +8 is 15, +16 is 31. You can't continue the problem because there are only 26 letters in the alphabet.
New contributor
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The property seems to be related to:
Binary equivalents of the symbols/alphabets etc.
Explanation:
Binary equivalents for the following can be written as:
A -- 01000001
B -- 01000010
D -- 01000100
H -- 01001000
P -- 01010000
0 -- 00110000
( -- 00101000
So the property is,
The sum of digits in the binary equivalents is two
Or
The binary equivalents of all these have two
1
s and six0
s.
Looking at the binary equivalents of the alphabets, one can see that no other alphabets share this property
The title (Thanks to @trolley813):
Two might refer to the sum of the digits, which is indeed two!
Title might mean that the sum [of digits in] letters is not different from two [but is equal to two]
Old (and wrong) answer
The property is:
The index if each alphabet is equal to the sum of indices of the preceding alphabets in the sequence +1.
And,
There is no other alphabet with the index
1+2+4+8+16+1 = 32
$endgroup$
1
$begingroup$
You've got it. Binary was what I was looking for but there are some other interesting patterns came out of this puzzle. I have added another hint in case anyone wants to have a go without looking at your answer. For the same reason, I'll wait a couple of hours before I accept it. well done!
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
@Eagle Some refinement about the title, it probably should read "sum [of digits in] letters is not different from two [i.e. equals to 2]"
$endgroup$
– trolley813
yesterday
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
yesterday
$begingroup$
@trolley813 - Close but actually more of a play on words. Rot13(Fhz (Fbzr) yrggref ner abg gjb (gbb) qvssrerag) with Rot13(Fhz) and Rot13(gjb) giving clues to what I was looking for ;-)
$endgroup$
– ElPedro
yesterday
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
The property seems to be related to:
Binary equivalents of the symbols/alphabets etc.
Explanation:
Binary equivalents for the following can be written as:
A -- 01000001
B -- 01000010
D -- 01000100
H -- 01001000
P -- 01010000
0 -- 00110000
( -- 00101000
So the property is,
The sum of digits in the binary equivalents is two
Or
The binary equivalents of all these have two
1
s and six0
s.
Looking at the binary equivalents of the alphabets, one can see that no other alphabets share this property
The title (Thanks to @trolley813):
Two might refer to the sum of the digits, which is indeed two!
Title might mean that the sum [of digits in] letters is not different from two [but is equal to two]
Old (and wrong) answer
The property is:
The index if each alphabet is equal to the sum of indices of the preceding alphabets in the sequence +1.
And,
There is no other alphabet with the index
1+2+4+8+16+1 = 32
$endgroup$
1
$begingroup$
You've got it. Binary was what I was looking for but there are some other interesting patterns came out of this puzzle. I have added another hint in case anyone wants to have a go without looking at your answer. For the same reason, I'll wait a couple of hours before I accept it. well done!
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
@Eagle Some refinement about the title, it probably should read "sum [of digits in] letters is not different from two [i.e. equals to 2]"
$endgroup$
– trolley813
yesterday
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
yesterday
$begingroup$
@trolley813 - Close but actually more of a play on words. Rot13(Fhz (Fbzr) yrggref ner abg gjb (gbb) qvssrerag) with Rot13(Fhz) and Rot13(gjb) giving clues to what I was looking for ;-)
$endgroup$
– ElPedro
yesterday
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
The property seems to be related to:
Binary equivalents of the symbols/alphabets etc.
Explanation:
Binary equivalents for the following can be written as:
A -- 01000001
B -- 01000010
D -- 01000100
H -- 01001000
P -- 01010000
0 -- 00110000
( -- 00101000
So the property is,
The sum of digits in the binary equivalents is two
Or
The binary equivalents of all these have two
1
s and six0
s.
Looking at the binary equivalents of the alphabets, one can see that no other alphabets share this property
The title (Thanks to @trolley813):
Two might refer to the sum of the digits, which is indeed two!
Title might mean that the sum [of digits in] letters is not different from two [but is equal to two]
Old (and wrong) answer
The property is:
The index if each alphabet is equal to the sum of indices of the preceding alphabets in the sequence +1.
And,
There is no other alphabet with the index
1+2+4+8+16+1 = 32
$endgroup$
The property seems to be related to:
Binary equivalents of the symbols/alphabets etc.
Explanation:
Binary equivalents for the following can be written as:
A -- 01000001
B -- 01000010
D -- 01000100
H -- 01001000
P -- 01010000
0 -- 00110000
( -- 00101000
So the property is,
The sum of digits in the binary equivalents is two
Or
The binary equivalents of all these have two
1
s and six0
s.
Looking at the binary equivalents of the alphabets, one can see that no other alphabets share this property
The title (Thanks to @trolley813):
Two might refer to the sum of the digits, which is indeed two!
Title might mean that the sum [of digits in] letters is not different from two [but is equal to two]
Old (and wrong) answer
The property is:
The index if each alphabet is equal to the sum of indices of the preceding alphabets in the sequence +1.
And,
There is no other alphabet with the index
1+2+4+8+16+1 = 32
edited yesterday
answered 2 days ago
EagleEagle
722226
722226
1
$begingroup$
You've got it. Binary was what I was looking for but there are some other interesting patterns came out of this puzzle. I have added another hint in case anyone wants to have a go without looking at your answer. For the same reason, I'll wait a couple of hours before I accept it. well done!
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
@Eagle Some refinement about the title, it probably should read "sum [of digits in] letters is not different from two [i.e. equals to 2]"
$endgroup$
– trolley813
yesterday
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
yesterday
$begingroup$
@trolley813 - Close but actually more of a play on words. Rot13(Fhz (Fbzr) yrggref ner abg gjb (gbb) qvssrerag) with Rot13(Fhz) and Rot13(gjb) giving clues to what I was looking for ;-)
$endgroup$
– ElPedro
yesterday
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
yesterday
add a comment |
1
$begingroup$
You've got it. Binary was what I was looking for but there are some other interesting patterns came out of this puzzle. I have added another hint in case anyone wants to have a go without looking at your answer. For the same reason, I'll wait a couple of hours before I accept it. well done!
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
@Eagle Some refinement about the title, it probably should read "sum [of digits in] letters is not different from two [i.e. equals to 2]"
$endgroup$
– trolley813
yesterday
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
yesterday
$begingroup$
@trolley813 - Close but actually more of a play on words. Rot13(Fhz (Fbzr) yrggref ner abg gjb (gbb) qvssrerag) with Rot13(Fhz) and Rot13(gjb) giving clues to what I was looking for ;-)
$endgroup$
– ElPedro
yesterday
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
yesterday
1
1
$begingroup$
You've got it. Binary was what I was looking for but there are some other interesting patterns came out of this puzzle. I have added another hint in case anyone wants to have a go without looking at your answer. For the same reason, I'll wait a couple of hours before I accept it. well done!
$endgroup$
– ElPedro
2 days ago
$begingroup$
You've got it. Binary was what I was looking for but there are some other interesting patterns came out of this puzzle. I have added another hint in case anyone wants to have a go without looking at your answer. For the same reason, I'll wait a couple of hours before I accept it. well done!
$endgroup$
– ElPedro
2 days ago
1
1
$begingroup$
@Eagle Some refinement about the title, it probably should read "sum [of digits in] letters is not different from two [i.e. equals to 2]"
$endgroup$
– trolley813
yesterday
$begingroup$
@Eagle Some refinement about the title, it probably should read "sum [of digits in] letters is not different from two [i.e. equals to 2]"
$endgroup$
– trolley813
yesterday
1
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
yesterday
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
yesterday
$begingroup$
@trolley813 - Close but actually more of a play on words. Rot13(Fhz (Fbzr) yrggref ner abg gjb (gbb) qvssrerag) with Rot13(Fhz) and Rot13(gjb) giving clues to what I was looking for ;-)
$endgroup$
– ElPedro
yesterday
$begingroup$
@trolley813 - Close but actually more of a play on words. Rot13(Fhz (Fbzr) yrggref ner abg gjb (gbb) qvssrerag) with Rot13(Fhz) and Rot13(gjb) giving clues to what I was looking for ;-)
$endgroup$
– ElPedro
yesterday
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
yesterday
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
$endgroup$
2
$begingroup$
That wasn't what I was looking for but is indeed true and is possibly a side effect of the answer that I was looking for so +1 but I won't mark it as accepted yet.
$endgroup$
– ElPedro
2 days ago
2
$begingroup$
@ElPedro: It is a side effect, but it hinges on rot13(jurer gur nycunorg fgnegf ba gur NFPVV gnoyr. Vtaber gur svefg ovg bs gur NFPVV inyhr (orpnhfr vg vf nyjnlf 1 naq arire punatrf sebz N gb M), ohg QB erzrzore gung vg nyjnlf pbagevohgrf gb bar bs gur gjb 1-ovgf va lbhe vagraqrq nafjre. Vtabevat gur svefg ovg, N rssrpgviryl fgnegf ng ahzrevpny inyhr 1, naq rirel "cbjre bs gjb" yrggre nqqf rknpgyl 1 1-ovg gb gur pbhag, juvpu rkcynvaf jul obgu nafjref ner pbeerpg. Vs N unq fgnegrq ba n qvssrerag NFPVV inyhr, vg znl abg unir orra gur pnfr.)
$endgroup$
– Flater
yesterday
$begingroup$
@Flater - Thanks for the explanation. It makes complete sense. As I said, some pretty interesting things have come out of what I at first though was a pretty simple puzzle :)
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
$endgroup$
2
$begingroup$
That wasn't what I was looking for but is indeed true and is possibly a side effect of the answer that I was looking for so +1 but I won't mark it as accepted yet.
$endgroup$
– ElPedro
2 days ago
2
$begingroup$
@ElPedro: It is a side effect, but it hinges on rot13(jurer gur nycunorg fgnegf ba gur NFPVV gnoyr. Vtaber gur svefg ovg bs gur NFPVV inyhr (orpnhfr vg vf nyjnlf 1 naq arire punatrf sebz N gb M), ohg QB erzrzore gung vg nyjnlf pbagevohgrf gb bar bs gur gjb 1-ovgf va lbhe vagraqrq nafjre. Vtabevat gur svefg ovg, N rssrpgviryl fgnegf ng ahzrevpny inyhr 1, naq rirel "cbjre bs gjb" yrggre nqqf rknpgyl 1 1-ovg gb gur pbhag, juvpu rkcynvaf jul obgu nafjref ner pbeerpg. Vs N unq fgnegrq ba n qvssrerag NFPVV inyhr, vg znl abg unir orra gur pnfr.)
$endgroup$
– Flater
yesterday
$begingroup$
@Flater - Thanks for the explanation. It makes complete sense. As I said, some pretty interesting things have come out of what I at first though was a pretty simple puzzle :)
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
$endgroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
answered 2 days ago
Deusovi♦Deusovi
63.3k6216273
63.3k6216273
2
$begingroup$
That wasn't what I was looking for but is indeed true and is possibly a side effect of the answer that I was looking for so +1 but I won't mark it as accepted yet.
$endgroup$
– ElPedro
2 days ago
2
$begingroup$
@ElPedro: It is a side effect, but it hinges on rot13(jurer gur nycunorg fgnegf ba gur NFPVV gnoyr. Vtaber gur svefg ovg bs gur NFPVV inyhr (orpnhfr vg vf nyjnlf 1 naq arire punatrf sebz N gb M), ohg QB erzrzore gung vg nyjnlf pbagevohgrf gb bar bs gur gjb 1-ovgf va lbhe vagraqrq nafjre. Vtabevat gur svefg ovg, N rssrpgviryl fgnegf ng ahzrevpny inyhr 1, naq rirel "cbjre bs gjb" yrggre nqqf rknpgyl 1 1-ovg gb gur pbhag, juvpu rkcynvaf jul obgu nafjref ner pbeerpg. Vs N unq fgnegrq ba n qvssrerag NFPVV inyhr, vg znl abg unir orra gur pnfr.)
$endgroup$
– Flater
yesterday
$begingroup$
@Flater - Thanks for the explanation. It makes complete sense. As I said, some pretty interesting things have come out of what I at first though was a pretty simple puzzle :)
$endgroup$
– ElPedro
yesterday
add a comment |
2
$begingroup$
That wasn't what I was looking for but is indeed true and is possibly a side effect of the answer that I was looking for so +1 but I won't mark it as accepted yet.
$endgroup$
– ElPedro
2 days ago
2
$begingroup$
@ElPedro: It is a side effect, but it hinges on rot13(jurer gur nycunorg fgnegf ba gur NFPVV gnoyr. Vtaber gur svefg ovg bs gur NFPVV inyhr (orpnhfr vg vf nyjnlf 1 naq arire punatrf sebz N gb M), ohg QB erzrzore gung vg nyjnlf pbagevohgrf gb bar bs gur gjb 1-ovgf va lbhe vagraqrq nafjre. Vtabevat gur svefg ovg, N rssrpgviryl fgnegf ng ahzrevpny inyhr 1, naq rirel "cbjre bs gjb" yrggre nqqf rknpgyl 1 1-ovg gb gur pbhag, juvpu rkcynvaf jul obgu nafjref ner pbeerpg. Vs N unq fgnegrq ba n qvssrerag NFPVV inyhr, vg znl abg unir orra gur pnfr.)
$endgroup$
– Flater
yesterday
$begingroup$
@Flater - Thanks for the explanation. It makes complete sense. As I said, some pretty interesting things have come out of what I at first though was a pretty simple puzzle :)
$endgroup$
– ElPedro
yesterday
2
2
$begingroup$
That wasn't what I was looking for but is indeed true and is possibly a side effect of the answer that I was looking for so +1 but I won't mark it as accepted yet.
$endgroup$
– ElPedro
2 days ago
$begingroup$
That wasn't what I was looking for but is indeed true and is possibly a side effect of the answer that I was looking for so +1 but I won't mark it as accepted yet.
$endgroup$
– ElPedro
2 days ago
2
2
$begingroup$
@ElPedro: It is a side effect, but it hinges on rot13(jurer gur nycunorg fgnegf ba gur NFPVV gnoyr. Vtaber gur svefg ovg bs gur NFPVV inyhr (orpnhfr vg vf nyjnlf 1 naq arire punatrf sebz N gb M), ohg QB erzrzore gung vg nyjnlf pbagevohgrf gb bar bs gur gjb 1-ovgf va lbhe vagraqrq nafjre. Vtabevat gur svefg ovg, N rssrpgviryl fgnegf ng ahzrevpny inyhr 1, naq rirel "cbjre bs gjb" yrggre nqqf rknpgyl 1 1-ovg gb gur pbhag, juvpu rkcynvaf jul obgu nafjref ner pbeerpg. Vs N unq fgnegrq ba n qvssrerag NFPVV inyhr, vg znl abg unir orra gur pnfr.)
$endgroup$
– Flater
yesterday
$begingroup$
@ElPedro: It is a side effect, but it hinges on rot13(jurer gur nycunorg fgnegf ba gur NFPVV gnoyr. Vtaber gur svefg ovg bs gur NFPVV inyhr (orpnhfr vg vf nyjnlf 1 naq arire punatrf sebz N gb M), ohg QB erzrzore gung vg nyjnlf pbagevohgrf gb bar bs gur gjb 1-ovgf va lbhe vagraqrq nafjre. Vtabevat gur svefg ovg, N rssrpgviryl fgnegf ng ahzrevpny inyhr 1, naq rirel "cbjre bs gjb" yrggre nqqf rknpgyl 1 1-ovg gb gur pbhag, juvpu rkcynvaf jul obgu nafjref ner pbeerpg. Vs N unq fgnegrq ba n qvssrerag NFPVV inyhr, vg znl abg unir orra gur pnfr.)
$endgroup$
– Flater
yesterday
$begingroup$
@Flater - Thanks for the explanation. It makes complete sense. As I said, some pretty interesting things have come out of what I at first though was a pretty simple puzzle :)
$endgroup$
– ElPedro
yesterday
$begingroup$
@Flater - Thanks for the explanation. It makes complete sense. As I said, some pretty interesting things have come out of what I at first though was a pretty simple puzzle :)
$endgroup$
– ElPedro
yesterday
add a comment |
$begingroup$
Is it
All the letters, symbols in this group have ASCII codes of form $2^m + 2^n$ where m and n are integers
Thus we have
From ascii code table,
$A = 65 = 64 + 1 = 2^6 + 2^0$
$B = 66 = 64 + 2 = 2^6 + 2^1$
$D = 68 = 64 + 4 = 2^6 + 2^2$
$H = 72 = 64 + 8 = 2^6 + 2^3$
$P = 80 = 64 + 16 = 2^6 + 2^4$
Other letters don't share this property because
64 + 32 = 96 which does not correspond to any letter. The letter
a
begins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
$endgroup$
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
2 days ago
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
2 days ago
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
I like the format of this answer more than the accepted one, simply because that's how I internally rephrased the accepted answer before even reading this one :) +1
$endgroup$
– Flater
yesterday
add a comment |
$begingroup$
Is it
All the letters, symbols in this group have ASCII codes of form $2^m + 2^n$ where m and n are integers
Thus we have
From ascii code table,
$A = 65 = 64 + 1 = 2^6 + 2^0$
$B = 66 = 64 + 2 = 2^6 + 2^1$
$D = 68 = 64 + 4 = 2^6 + 2^2$
$H = 72 = 64 + 8 = 2^6 + 2^3$
$P = 80 = 64 + 16 = 2^6 + 2^4$
Other letters don't share this property because
64 + 32 = 96 which does not correspond to any letter. The letter
a
begins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
$endgroup$
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
2 days ago
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
2 days ago
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
I like the format of this answer more than the accepted one, simply because that's how I internally rephrased the accepted answer before even reading this one :) +1
$endgroup$
– Flater
yesterday
add a comment |
$begingroup$
Is it
All the letters, symbols in this group have ASCII codes of form $2^m + 2^n$ where m and n are integers
Thus we have
From ascii code table,
$A = 65 = 64 + 1 = 2^6 + 2^0$
$B = 66 = 64 + 2 = 2^6 + 2^1$
$D = 68 = 64 + 4 = 2^6 + 2^2$
$H = 72 = 64 + 8 = 2^6 + 2^3$
$P = 80 = 64 + 16 = 2^6 + 2^4$
Other letters don't share this property because
64 + 32 = 96 which does not correspond to any letter. The letter
a
begins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
$endgroup$
Is it
All the letters, symbols in this group have ASCII codes of form $2^m + 2^n$ where m and n are integers
Thus we have
From ascii code table,
$A = 65 = 64 + 1 = 2^6 + 2^0$
$B = 66 = 64 + 2 = 2^6 + 2^1$
$D = 68 = 64 + 4 = 2^6 + 2^2$
$H = 72 = 64 + 8 = 2^6 + 2^3$
$P = 80 = 64 + 16 = 2^6 + 2^4$
Other letters don't share this property because
64 + 32 = 96 which does not correspond to any letter. The letter
a
begins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
edited 2 days ago
Eagle
722226
722226
answered 2 days ago
RedBaronRedBaron
42636
42636
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
2 days ago
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
2 days ago
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
I like the format of this answer more than the accepted one, simply because that's how I internally rephrased the accepted answer before even reading this one :) +1
$endgroup$
– Flater
yesterday
add a comment |
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
2 days ago
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
2 days ago
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
I like the format of this answer more than the accepted one, simply because that's how I internally rephrased the accepted answer before even reading this one :) +1
$endgroup$
– Flater
yesterday
1
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
2 days ago
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
2 days ago
3
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
2 days ago
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
2 days ago
1
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
2 days ago
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
2 days ago
1
1
$begingroup$
I like the format of this answer more than the accepted one, simply because that's how I internally rephrased the accepted answer before even reading this one :) +1
$endgroup$
– Flater
yesterday
$begingroup$
I like the format of this answer more than the accepted one, simply because that's how I internally rephrased the accepted answer before even reading this one :) +1
$endgroup$
– Flater
yesterday
add a comment |
$begingroup$
I think it's:
Each letter's alphanumeric value is double its predecessor, which also means, sum two times the alphanumeric value of the previous letter
This means that:
Starting from A=1 we get the sequence 1,2,4,8,16,... which corresponds to the sequence A,B,D,H,P
New contributor
$endgroup$
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
2 days ago
add a comment |
$begingroup$
I think it's:
Each letter's alphanumeric value is double its predecessor, which also means, sum two times the alphanumeric value of the previous letter
This means that:
Starting from A=1 we get the sequence 1,2,4,8,16,... which corresponds to the sequence A,B,D,H,P
New contributor
$endgroup$
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
2 days ago
add a comment |
$begingroup$
I think it's:
Each letter's alphanumeric value is double its predecessor, which also means, sum two times the alphanumeric value of the previous letter
This means that:
Starting from A=1 we get the sequence 1,2,4,8,16,... which corresponds to the sequence A,B,D,H,P
New contributor
$endgroup$
I think it's:
Each letter's alphanumeric value is double its predecessor, which also means, sum two times the alphanumeric value of the previous letter
This means that:
Starting from A=1 we get the sequence 1,2,4,8,16,... which corresponds to the sequence A,B,D,H,P
New contributor
New contributor
answered 2 days ago
SaeleasSaeleas
1612
1612
New contributor
New contributor
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
2 days ago
add a comment |
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
2 days ago
1
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
2 days ago
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
2 days ago
add a comment |
$begingroup$
Each time you add the position of the letter (A is 1 and B is 2), the next letter's position is the sum of the previous +1.
Thus, 1+2 is 3, +4 is 7, +8 is 15, +16 is 31. You can't continue the problem because there are only 26 letters in the alphabet.
New contributor
$endgroup$
add a comment |
$begingroup$
Each time you add the position of the letter (A is 1 and B is 2), the next letter's position is the sum of the previous +1.
Thus, 1+2 is 3, +4 is 7, +8 is 15, +16 is 31. You can't continue the problem because there are only 26 letters in the alphabet.
New contributor
$endgroup$
add a comment |
$begingroup$
Each time you add the position of the letter (A is 1 and B is 2), the next letter's position is the sum of the previous +1.
Thus, 1+2 is 3, +4 is 7, +8 is 15, +16 is 31. You can't continue the problem because there are only 26 letters in the alphabet.
New contributor
$endgroup$
Each time you add the position of the letter (A is 1 and B is 2), the next letter's position is the sum of the previous +1.
Thus, 1+2 is 3, +4 is 7, +8 is 15, +16 is 31. You can't continue the problem because there are only 26 letters in the alphabet.
New contributor
New contributor
answered 2 days ago
CStafford-14CStafford-14
16510
16510
New contributor
New contributor
add a comment |
add a comment |
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$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
2 days ago
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
2 days ago
1
$begingroup$
Does a
!
also share this property?$endgroup$
– Eagle
2 days ago
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
2 days ago