Sum letters are not two different
$begingroup$
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
0and 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
asked Apr 18 at 8:22
ElPedroElPedro
303210
$endgroup$
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
0and 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
asked Apr 18 at 8:22
ElPedroElPedro
303210
$endgroup$
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
Apr 18 at 9:18
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
Apr 18 at 9:19
1
$begingroup$
Does a!also share this property?
$endgroup$
– Eagle
Apr 18 at 12:04
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
Apr 18 at 12:06
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
0and 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
asked Apr 18 at 8:22
ElPedroElPedro
303210
$endgroup$
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
0and 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
asked Apr 18 at 8:22
ElPedroElPedro
303210
asked Apr 18 at 8:22
ElPedroElPedro
303210
edited Apr 18 at 12:23
ElPedro
asked Apr 18 at 8:22
ElPedroElPedro
303210
asked Apr 18 at 8:22
ElPedroElPedro
303210
asked Apr 18 at 8:22
ElPedroElPedro
303210
303210
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
Apr 18 at 9:18
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
Apr 18 at 9:19
1
$begingroup$
Does a!also share this property?
$endgroup$
– Eagle
Apr 18 at 12:04
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
Apr 18 at 12:06
add a comment |
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
Apr 18 at 9:18
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
Apr 18 at 9:19
1
$begingroup$
Does a!also share this property?
$endgroup$
– Eagle
Apr 18 at 12:04
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
Apr 18 at 12:06
5
5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
Apr 18 at 9:18
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
Apr 18 at 9:18
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
Apr 18 at 9:19
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
Apr 18 at 9:19
1
1
$begingroup$
Does a
! also share this property?$endgroup$
– Eagle
Apr 18 at 12:04
$begingroup$
Does a
! also share this property?$endgroup$
– Eagle
Apr 18 at 12:04
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
Apr 18 at 12:06
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
Apr 18 at 12:06
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
1s and six0s.
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
answered Apr 18 at 9:49
EagleEagle
718226
$endgroup$
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
Apr 18 at 12:26
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
Apr 19 at 10:03
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
Apr 19 at 11:15
$begingroup$
@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
Apr 19 at 14:51
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
Apr 19 at 14:58
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
answered Apr 18 at 9:20
Deusovi♦Deusovi
63.3k6216273
$endgroup$
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
Apr 18 at 9:23
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
Apr 19 at 8:21
$begingroup$
@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
Apr 19 at 14:43
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
abegins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
edited Apr 18 at 19:50
Eagle
718226
answered Apr 18 at 12:15
RedBaronRedBaron
42636
$endgroup$
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
Apr 18 at 12:20
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
Apr 18 at 12:22
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
Apr 18 at 12:42
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
Apr 19 at 8:24
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
answered Apr 18 at 9:58
SaeleasSaeleas
1612
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
Apr 18 at 12:14
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.
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
1s and six0s.
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
answered Apr 18 at 9:49
EagleEagle
718226
$endgroup$
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
Apr 18 at 12:26
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
Apr 19 at 10:03
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
Apr 19 at 11:15
$begingroup$
@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
Apr 19 at 14:51
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
Apr 19 at 14:58
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
1s and six0s.
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
answered Apr 18 at 9:49
EagleEagle
718226
$endgroup$
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
Apr 18 at 12:26
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
Apr 19 at 10:03
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
Apr 19 at 11:15
$begingroup$
@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
Apr 19 at 14:51
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
Apr 19 at 14:58
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
1s and six0s.
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
answered Apr 18 at 9:49
EagleEagle
718226
$endgroup$
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
1s and six0s.
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
answered Apr 18 at 9:49
EagleEagle
718226
edited Apr 19 at 11:15
answered Apr 18 at 9:49
EagleEagle
718226
answered Apr 18 at 9:49
EagleEagle
718226
answered Apr 18 at 9:49
EagleEagle
718226
718226
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
Apr 18 at 12:26
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
Apr 19 at 10:03
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
Apr 19 at 11:15
$begingroup$
@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
Apr 19 at 14:51
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
Apr 19 at 14:58
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
Apr 18 at 12:26
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
Apr 19 at 10:03
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
Apr 19 at 11:15
$begingroup$
@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
Apr 19 at 14:51
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
Apr 19 at 14:58
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
Apr 18 at 12:26
$begingroup$
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
Apr 18 at 12:26
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
Apr 19 at 10:03
$begingroup$
@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
Apr 19 at 10:03
1
1
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
Apr 19 at 11:15
$begingroup$
Thanks a lot @trolley813 ! I've edited it
$endgroup$
– Eagle
Apr 19 at 11:15
$begingroup$
@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
Apr 19 at 14:51
$begingroup$
@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
Apr 19 at 14:51
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
Apr 19 at 14:58
$begingroup$
A bit contrived, I know, but left room for a couple of hints.
$endgroup$
– ElPedro
Apr 19 at 14:58
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
answered Apr 18 at 9:20
Deusovi♦Deusovi
63.3k6216273
$endgroup$
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
Apr 18 at 9:23
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
Apr 19 at 8:21
$begingroup$
@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
Apr 19 at 14:43
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
answered Apr 18 at 9:20
Deusovi♦Deusovi
63.3k6216273
$endgroup$
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
Apr 18 at 9:23
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
Apr 19 at 8:21
$begingroup$
@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
Apr 19 at 14:43
add a comment |
$begingroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
answered Apr 18 at 9:20
Deusovi♦Deusovi
63.3k6216273
$endgroup$
The property is that
each of their alphanumeric values (A=1, B=2, C=3...) is a power of 2.
answered Apr 18 at 9:20
Deusovi♦Deusovi
63.3k6216273
answered Apr 18 at 9:20
Deusovi♦Deusovi
63.3k6216273
answered Apr 18 at 9:20
Deusovi♦Deusovi
63.3k6216273
answered Apr 18 at 9:20
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
Apr 18 at 9:23
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
Apr 19 at 8:21
$begingroup$
@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
Apr 19 at 14:43
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
Apr 18 at 9:23
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
Apr 19 at 8:21
$begingroup$
@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
Apr 19 at 14:43
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
Apr 18 at 9:23
$begingroup$
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
Apr 18 at 9:23
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
Apr 19 at 8:21
$begingroup$
@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
Apr 19 at 8:21
$begingroup$
@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
Apr 19 at 14:43
$begingroup$
@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
Apr 19 at 14:43
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
abegins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
edited Apr 18 at 19:50
Eagle
718226
answered Apr 18 at 12:15
RedBaronRedBaron
42636
$endgroup$
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
Apr 18 at 12:20
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
Apr 18 at 12:22
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
Apr 18 at 12:42
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
Apr 19 at 8:24
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
abegins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
edited Apr 18 at 19:50
Eagle
718226
answered Apr 18 at 12:15
RedBaronRedBaron
42636
$endgroup$
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
Apr 18 at 12:20
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
Apr 18 at 12:22
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
Apr 18 at 12:42
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
Apr 19 at 8:24
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
abegins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
edited Apr 18 at 19:50
Eagle
718226
answered Apr 18 at 12:15
RedBaronRedBaron
42636
$endgroup$
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
abegins at 97
For the newer hints
$0 = 48 = 32 + 16 = 2^5 + 2^4$
$( = 40 = 32 + 8 = 2^5 + 2^3$
edited Apr 18 at 19:50
Eagle
718226
answered Apr 18 at 12:15
RedBaronRedBaron
42636
edited Apr 18 at 19:50
Eagle
718226
edited Apr 18 at 19:50
Eagle
718226
edited Apr 18 at 19:50
Eagle
718226
718226
answered Apr 18 at 12:15
RedBaronRedBaron
42636
answered Apr 18 at 12:15
RedBaronRedBaron
42636
answered Apr 18 at 12:15
RedBaronRedBaron
42636
42636
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
Apr 18 at 12:20
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
Apr 18 at 12:22
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
Apr 18 at 12:42
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
Apr 19 at 8:24
add a comment |
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
Apr 18 at 12:20
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
Apr 18 at 12:22
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
Apr 18 at 12:42
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
Apr 19 at 8:24
1
1
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
Apr 18 at 12:20
$begingroup$
Obviously moving in the right direction with the ASCII table.
$endgroup$
– ElPedro
Apr 18 at 12:20
3
3
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
Apr 18 at 12:22
$begingroup$
@ElPedro I guess Akari has formalized the informal property of my answer much better in his answer
$endgroup$
– RedBaron
Apr 18 at 12:22
1
1
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
Apr 18 at 12:42
$begingroup$
Still a good answer though :)
$endgroup$
– ElPedro
Apr 18 at 12:42
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
Apr 19 at 8:24
$begingroup$
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
Apr 19 at 8:24
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
answered Apr 18 at 9:58
SaeleasSaeleas
1612
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
Apr 18 at 12:14
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
answered Apr 18 at 9:58
SaeleasSaeleas
1612
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
Apr 18 at 12:14
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
answered Apr 18 at 9:58
SaeleasSaeleas
1612
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
answered Apr 18 at 9:58
SaeleasSaeleas
1612
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 18 at 9:58
SaeleasSaeleas
1612
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 18 at 9:58
SaeleasSaeleas
1612
answered Apr 18 at 9:58
SaeleasSaeleas
1612
1612
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Saeleas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
Apr 18 at 12:14
add a comment |
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
Apr 18 at 12:14
1
1
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
Apr 18 at 12:14
$begingroup$
Welcome! Again, a great answer but not exactly what I am looking for. +1 all the same.
$endgroup$
– ElPedro
Apr 18 at 12:14
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.
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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.
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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.
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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.
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
answered Apr 18 at 11:41
CStafford-14CStafford-14
30710
30710
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CStafford-14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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5
$begingroup$
Are you sure B shouldn't be included too?
$endgroup$
– Deusovi♦
Apr 18 at 9:18
$begingroup$
@Deusovi You are indeed correct. I had missed that. I'll update the question. Thanks.
$endgroup$
– ElPedro
Apr 18 at 9:19
1
$begingroup$
Does a
!also share this property?$endgroup$
– Eagle
Apr 18 at 12:04
$begingroup$
@Akari Yes it does. I have not listed them all.
$endgroup$
– ElPedro
Apr 18 at 12:06