3 Solutions to the Liber CCXX II:76 Riddle
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**3 Solutions to the Liber CCXX II:76 Riddle
**http://img403.imageshack.us/img403/4237/ii76.png
**Solution 1
**
The sequence opens with"4 6 3 8"
4 times 6 is 24
3 times 8 is 24The second row opens with
"x 24 89"
"x" is the 24th letter of the alphabet, and 89 is the 24th prime.
So the key is the English alphabet, in alphabetical order, attached to the prime numbers.
The first 26 primes
1 A 2
2 B 3
3 C 5
4 D 7
5 E 11
6 F 13
7 G 17
8 H 19
9 I 23
10 J 29
11 K 31
12 L 37
13 M 41
14 N 43
15 O 47
16 P 53
17 Q 59
18 R 61
19 S 67
20 T 71
21 U 73
22 V 79
23 W 83
24 X 89
25 Y 97
26 Z 101So, on to deciphering.
The first line is
4638ABK24aLGMOR3YIf we take the first two pairs of numbers as each representing 24 (based on our initial observation that 4 times 6 is 24, and 3 times 8 is 24), then we get
24 24 ABK 24 aLGMOR 3 Y
Using our prime values for the letters, we get
ABK = 2+3+31 = 36
aLGMOR = 2+37+17+41+53+61 = 211
Y = 97Adding it all together, we get:
24+24+36+24+211+3+97 = 419The second line is
x 24 89 RPSTOVAL.If we ignore the “x 24 89” as the cipher, we get
RPSTOVAL = 71+59+47+23+53+13+2+37 = 417
The final answer is then literally and metaphorically “between the lines” – the average between lines 1 and 2 is 418! Abrahadabra!
**Solution 2
**
If we use standard Hebrew gematria on the second row letters, we get:RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456
On the first row, if we ignore the 4638 and 24 as being part of the embedded cipher clue, we have
ABK, aLGMOR, 3 and Y
Using standard Hebrew gematria on row 2, as we did in Solution 2, we get
23, 344, 3, and 10, which add up to 380
380 + 38 = 418
456 – 38 = 418In other words, once again the average of rows 1 and 2 are 418!
Further, 38 is one of the numbers in the code!
**Solution 3
**
So, what happens if we take the Hebrew alphabet, and line them up to the first 22 primes?2 aleph
3 beth
5 gimel
7 daleth
11 Heh
13 waw
17 zayin
19 cheth
23 theth
29 Yod
31 Kaph
37 Lamed
41 Mem
43 Nun
47 Samek
53 Ayin
59 Peh
61 Tzaddi
67 Qoph
71 Resh
73 Shin
79 TavUsing this cipher, we get the following values for the letters of the first row.
ABK = Aleph-Beth-Kaph = 2+3+31 = 36
ALGMOR = aleph-lamed-gimel-mem-ayin-resh = 2+37+5+41+53+61+29 = 199
3Y = 3 yod = 3 times 29 = 8736+199+87 = 322
That leaves the 4638 and the 24. If we treat the 46 and 38 as ciphers for 24, we get
322+24+24+24 = 394
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 71+59+47+23+53+13+2+37 = 305
305 + 23 + 89 = 418!
Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…
418!
So we have one row, the total of all characters being 418 - 24, and another row where all characters total 418 + 24!
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**3 Solutions to the Liber CCXX II:76 Riddle
**http://img403.imageshack.us/img403/4237/ii76.png
**Solution 1
**
The sequence opens with"4 6 3 8"
4 times 6 is 24
3 times 8 is 24The second row opens with
"x 24 89"
"x" is the 24th letter of the alphabet, and 89 is the 24th prime.
So the key is the English alphabet, in alphabetical order, attached to the prime numbers.
The first 26 primes
1 A 2
2 B 3
3 C 5
4 D 7
5 E 11
6 F 13
7 G 17
8 H 19
9 I 23
10 J 29
11 K 31
12 L 37
13 M 41
14 N 43
15 O 47
16 P 53
17 Q 59
18 R 61
19 S 67
20 T 71
21 U 73
22 V 79
23 W 83
24 X 89
25 Y 97
26 Z 101So, on to deciphering.
The first line is
4638ABK24aLGMOR3YIf we take the first two pairs of numbers as each representing 24 (based on our initial observation that 4 times 6 is 24, and 3 times 8 is 24), then we get
24 24 ABK 24 aLGMOR 3 Y
Using our prime values for the letters, we get
ABK = 2+3+31 = 36
aLGMOR = 2+37+17+41+53+61 = 211
Y = 97Adding it all together, we get:
24+24+36+24+211+3+97 = 419The second line is
x 24 89 RPSTOVAL.If we ignore the “x 24 89” as the cipher, we get
RPSTOVAL = 71+59+47+23+53+13+2+37 = 417
The final answer is then literally and metaphorically “between the lines” – the average between lines 1 and 2 is 418! Abrahadabra!
**Solution 2
**
If we use standard Hebrew gematria on the second row letters, we get:RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456
On the first row, if we ignore the 4638 and 24 as being part of the embedded cipher clue, we have
ABK, aLGMOR, 3 and Y
Using standard Hebrew gematria on row 2, as we did in Solution 2, we get
23, 344, 3, and 10, which add up to 380
380 + 38 = 418
456 – 38 = 418In other words, once again the average of rows 1 and 2 are 418!
Further, 38 is one of the numbers in the code!
**Solution 3
**
So, what happens if we take the Hebrew alphabet, and line them up to the first 22 primes?2 aleph
3 beth
5 gimel
7 daleth
11 Heh
13 waw
17 zayin
19 cheth
23 theth
29 Yod
31 Kaph
37 Lamed
41 Mem
43 Nun
47 Samek
53 Ayin
59 Peh
61 Tzaddi
67 Qoph
71 Resh
73 Shin
79 TavUsing this cipher, we get the following values for the letters of the first row.
ABK = Aleph-Beth-Kaph = 2+3+31 = 36
ALGMOR = aleph-lamed-gimel-mem-ayin-resh = 2+37+5+41+53+61+29 = 199
3Y = 3 yod = 3 times 29 = 8736+199+87 = 322
That leaves the 4638 and the 24. If we treat the 46 and 38 as ciphers for 24, we get
322+24+24+24 = 394
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 71+59+47+23+53+13+2+37 = 305
305 + 23 + 89 = 418!
Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…
418!
So we have one row, the total of all characters being 418 - 24, and another row where all characters total 418 + 24!
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**3 Solutions to the Liber CCXX II:76 Riddle
**http://img403.imageshack.us/img403/4237/ii76.png
**Solution 1
**
The sequence opens with"4 6 3 8"
4 times 6 is 24
3 times 8 is 24The second row opens with
"x 24 89"
"x" is the 24th letter of the alphabet, and 89 is the 24th prime.
So the key is the English alphabet, in alphabetical order, attached to the prime numbers.
The first 26 primes
1 A 2
2 B 3
3 C 5
4 D 7
5 E 11
6 F 13
7 G 17
8 H 19
9 I 23
10 J 29
11 K 31
12 L 37
13 M 41
14 N 43
15 O 47
16 P 53
17 Q 59
18 R 61
19 S 67
20 T 71
21 U 73
22 V 79
23 W 83
24 X 89
25 Y 97
26 Z 101So, on to deciphering.
The first line is
4638ABK24aLGMOR3YIf we take the first two pairs of numbers as each representing 24 (based on our initial observation that 4 times 6 is 24, and 3 times 8 is 24), then we get
24 24 ABK 24 aLGMOR 3 Y
Using our prime values for the letters, we get
ABK = 2+3+31 = 36
aLGMOR = 2+37+17+41+53+61 = 211
Y = 97Adding it all together, we get:
24+24+36+24+211+3+97 = 419The second line is
x 24 89 RPSTOVAL.If we ignore the “x 24 89” as the cipher, we get
RPSTOVAL = 71+59+47+23+53+13+2+37 = 417
The final answer is then literally and metaphorically “between the lines” – the average between lines 1 and 2 is 418! Abrahadabra!
**Solution 2
**
If we use standard Hebrew gematria on the second row letters, we get:RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456
On the first row, if we ignore the 4638 and 24 as being part of the embedded cipher clue, we have
ABK, aLGMOR, 3 and Y
Using standard Hebrew gematria on row 2, as we did in Solution 2, we get
23, 344, 3, and 10, which add up to 380
380 + 38 = 418
456 – 38 = 418In other words, once again the average of rows 1 and 2 are 418!
Further, 38 is one of the numbers in the code!
**Solution 3
**
So, what happens if we take the Hebrew alphabet, and line them up to the first 22 primes?2 aleph
3 beth
5 gimel
7 daleth
11 Heh
13 waw
17 zayin
19 cheth
23 theth
29 Yod
31 Kaph
37 Lamed
41 Mem
43 Nun
47 Samek
53 Ayin
59 Peh
61 Tzaddi
67 Qoph
71 Resh
73 Shin
79 TavUsing this cipher, we get the following values for the letters of the first row.
ABK = Aleph-Beth-Kaph = 2+3+31 = 36
ALGMOR = aleph-lamed-gimel-mem-ayin-resh = 2+37+5+41+53+61+29 = 199
3Y = 3 yod = 3 times 29 = 8736+199+87 = 322
That leaves the 4638 and the 24. If we treat the 46 and 38 as ciphers for 24, we get
322+24+24+24 = 394
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 71+59+47+23+53+13+2+37 = 305
305 + 23 + 89 = 418!
Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…
418!
So we have one row, the total of all characters being 418 - 24, and another row where all characters total 418 + 24!
Well, the x 24 89 prime number bit, I had mulling around in my head, and have been playing with for a while.
But, you'll notice that solutions 1 and 2 require only adding a subset of the contents of the line. So I think it would be fair to say that those two solutions involve "cherry picking" to get to 418.
Solution 3 is the only one of the 3 that uses all the letters and numbers of each row.
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**3 Solutions to the Liber CCXX II:76 Riddle
**http://img403.imageshack.us/img403/4237/ii76.png
**Solution 1
**
The sequence opens with"4 6 3 8"
4 times 6 is 24
3 times 8 is 24The second row opens with
"x 24 89"
"x" is the 24th letter of the alphabet, and 89 is the 24th prime.
So the key is the English alphabet, in alphabetical order, attached to the prime numbers.
The first 26 primes
1 A 2
2 B 3
3 C 5
4 D 7
5 E 11
6 F 13
7 G 17
8 H 19
9 I 23
10 J 29
11 K 31
12 L 37
13 M 41
14 N 43
15 O 47
16 P 53
17 Q 59
18 R 61
19 S 67
20 T 71
21 U 73
22 V 79
23 W 83
24 X 89
25 Y 97
26 Z 101So, on to deciphering.
The first line is
4638ABK24aLGMOR3YIf we take the first two pairs of numbers as each representing 24 (based on our initial observation that 4 times 6 is 24, and 3 times 8 is 24), then we get
24 24 ABK 24 aLGMOR 3 Y
Using our prime values for the letters, we get
ABK = 2+3+31 = 36
aLGMOR = 2+37+17+41+53+61 = 211
Y = 97Adding it all together, we get:
24+24+36+24+211+3+97 = 419The second line is
x 24 89 RPSTOVAL.If we ignore the “x 24 89” as the cipher, we get
RPSTOVAL = 71+59+47+23+53+13+2+37 = 417
The final answer is then literally and metaphorically “between the lines” – the average between lines 1 and 2 is 418! Abrahadabra!
**Solution 2
**
If we use standard Hebrew gematria on the second row letters, we get:RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456
On the first row, if we ignore the 4638 and 24 as being part of the embedded cipher clue, we have
ABK, aLGMOR, 3 and Y
Using standard Hebrew gematria on row 2, as we did in Solution 2, we get
23, 344, 3, and 10, which add up to 380
380 + 38 = 418
456 – 38 = 418In other words, once again the average of rows 1 and 2 are 418!
Further, 38 is one of the numbers in the code!
**Solution 3
**
So, what happens if we take the Hebrew alphabet, and line them up to the first 22 primes?2 aleph
3 beth
5 gimel
7 daleth
11 Heh
13 waw
17 zayin
19 cheth
23 theth
29 Yod
31 Kaph
37 Lamed
41 Mem
43 Nun
47 Samek
53 Ayin
59 Peh
61 Tzaddi
67 Qoph
71 Resh
73 Shin
79 TavUsing this cipher, we get the following values for the letters of the first row.
ABK = Aleph-Beth-Kaph = 2+3+31 = 36
ALGMOR = aleph-lamed-gimel-mem-ayin-resh = 2+37+5+41+53+61+29 = 199
3Y = 3 yod = 3 times 29 = 8736+199+87 = 322
That leaves the 4638 and the 24. If we treat the 46 and 38 as ciphers for 24, we get
322+24+24+24 = 394
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 71+59+47+23+53+13+2+37 = 305
305 + 23 + 89 = 418!
Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…
418!
So we have one row, the total of all characters being 418 - 24, and another row where all characters total 418 + 24!
"So - what's the glad word then Av?"
Abrahadabra!
"But remember, o chose none"
Oh... well in that case...
"Nothing is a secret key of this law. Sixty-one the Jews call it; I call it eight, eighty, four hundred & eighteen.
"To Μη - "nothing" - 418 (Pointed out in the Black Pearl V1 N2)
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**3 Solutions to the Liber CCXX II:76 Riddle
**http://img403.imageshack.us/img403/4237/ii76.png
**Solution 1
**
The sequence opens with"4 6 3 8"
4 times 6 is 24
3 times 8 is 24The second row opens with
"x 24 89"
"x" is the 24th letter of the alphabet, and 89 is the 24th prime.
So the key is the English alphabet, in alphabetical order, attached to the prime numbers.
The first 26 primes
1 A 2
2 B 3
3 C 5
4 D 7
5 E 11
6 F 13
7 G 17
8 H 19
9 I 23
10 J 29
11 K 31
12 L 37
13 M 41
14 N 43
15 O 47
16 P 53
17 Q 59
18 R 61
19 S 67
20 T 71
21 U 73
22 V 79
23 W 83
24 X 89
25 Y 97
26 Z 101So, on to deciphering.
The first line is
4638ABK24aLGMOR3YIf we take the first two pairs of numbers as each representing 24 (based on our initial observation that 4 times 6 is 24, and 3 times 8 is 24), then we get
24 24 ABK 24 aLGMOR 3 Y
Using our prime values for the letters, we get
ABK = 2+3+31 = 36
aLGMOR = 2+37+17+41+53+61 = 211
Y = 97Adding it all together, we get:
24+24+36+24+211+3+97 = 419The second line is
x 24 89 RPSTOVAL.If we ignore the “x 24 89” as the cipher, we get
RPSTOVAL = 71+59+47+23+53+13+2+37 = 417
The final answer is then literally and metaphorically “between the lines” – the average between lines 1 and 2 is 418! Abrahadabra!
**Solution 2
**
If we use standard Hebrew gematria on the second row letters, we get:RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456
On the first row, if we ignore the 4638 and 24 as being part of the embedded cipher clue, we have
ABK, aLGMOR, 3 and Y
Using standard Hebrew gematria on row 2, as we did in Solution 2, we get
23, 344, 3, and 10, which add up to 380
380 + 38 = 418
456 – 38 = 418In other words, once again the average of rows 1 and 2 are 418!
Further, 38 is one of the numbers in the code!
**Solution 3
**
So, what happens if we take the Hebrew alphabet, and line them up to the first 22 primes?2 aleph
3 beth
5 gimel
7 daleth
11 Heh
13 waw
17 zayin
19 cheth
23 theth
29 Yod
31 Kaph
37 Lamed
41 Mem
43 Nun
47 Samek
53 Ayin
59 Peh
61 Tzaddi
67 Qoph
71 Resh
73 Shin
79 TavUsing this cipher, we get the following values for the letters of the first row.
ABK = Aleph-Beth-Kaph = 2+3+31 = 36
ALGMOR = aleph-lamed-gimel-mem-ayin-resh = 2+37+5+41+53+61+29 = 199
3Y = 3 yod = 3 times 29 = 8736+199+87 = 322
That leaves the 4638 and the 24. If we treat the 46 and 38 as ciphers for 24, we get
322+24+24+24 = 394
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 71+59+47+23+53+13+2+37 = 305
305 + 23 + 89 = 418!
Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…
418!
So we have one row, the total of all characters being 418 - 24, and another row where all characters total 418 + 24!
That is very impressive!
"If we use standard Hebrew gematria on the second row letters, we get:
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456"
The number 456 is also "NVTH" (Nut, or Not) and Nu, Vav, Thav in full values = (224 +24 +418) = 666!!!
"Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…"
Actually the "X" could be put into the equation if you look at it as an archaic form of the letter "Thav" which is a an "X" (just a thought)
Thank you for sharing this, I have learned a great deal of things that I would not of thought in a million years!
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**3 Solutions to the Liber CCXX II:76 Riddle
**http://img403.imageshack.us/img403/4237/ii76.png
**Solution 1
**
The sequence opens with"4 6 3 8"
4 times 6 is 24
3 times 8 is 24The second row opens with
"x 24 89"
"x" is the 24th letter of the alphabet, and 89 is the 24th prime.
So the key is the English alphabet, in alphabetical order, attached to the prime numbers.
The first 26 primes
1 A 2
2 B 3
3 C 5
4 D 7
5 E 11
6 F 13
7 G 17
8 H 19
9 I 23
10 J 29
11 K 31
12 L 37
13 M 41
14 N 43
15 O 47
16 P 53
17 Q 59
18 R 61
19 S 67
20 T 71
21 U 73
22 V 79
23 W 83
24 X 89
25 Y 97
26 Z 101So, on to deciphering.
The first line is
4638ABK24aLGMOR3YIf we take the first two pairs of numbers as each representing 24 (based on our initial observation that 4 times 6 is 24, and 3 times 8 is 24), then we get
24 24 ABK 24 aLGMOR 3 Y
Using our prime values for the letters, we get
ABK = 2+3+31 = 36
aLGMOR = 2+37+17+41+53+61 = 211
Y = 97Adding it all together, we get:
24+24+36+24+211+3+97 = 419The second line is
x 24 89 RPSTOVAL.If we ignore the “x 24 89” as the cipher, we get
RPSTOVAL = 71+59+47+23+53+13+2+37 = 417
The final answer is then literally and metaphorically “between the lines” – the average between lines 1 and 2 is 418! Abrahadabra!
**Solution 2
**
If we use standard Hebrew gematria on the second row letters, we get:RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456
On the first row, if we ignore the 4638 and 24 as being part of the embedded cipher clue, we have
ABK, aLGMOR, 3 and Y
Using standard Hebrew gematria on row 2, as we did in Solution 2, we get
23, 344, 3, and 10, which add up to 380
380 + 38 = 418
456 – 38 = 418In other words, once again the average of rows 1 and 2 are 418!
Further, 38 is one of the numbers in the code!
**Solution 3
**
So, what happens if we take the Hebrew alphabet, and line them up to the first 22 primes?2 aleph
3 beth
5 gimel
7 daleth
11 Heh
13 waw
17 zayin
19 cheth
23 theth
29 Yod
31 Kaph
37 Lamed
41 Mem
43 Nun
47 Samek
53 Ayin
59 Peh
61 Tzaddi
67 Qoph
71 Resh
73 Shin
79 TavUsing this cipher, we get the following values for the letters of the first row.
ABK = Aleph-Beth-Kaph = 2+3+31 = 36
ALGMOR = aleph-lamed-gimel-mem-ayin-resh = 2+37+5+41+53+61+29 = 199
3Y = 3 yod = 3 times 29 = 8736+199+87 = 322
That leaves the 4638 and the 24. If we treat the 46 and 38 as ciphers for 24, we get
322+24+24+24 = 394
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 71+59+47+23+53+13+2+37 = 305
305 + 23 + 89 = 418!
Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…
418!
So we have one row, the total of all characters being 418 - 24, and another row where all characters total 418 + 24!
So....
I was playing around with the sequence of letters (transliterated into Hebrew), and I started to plug them into various translation tools, and put in different patterns of space marks to see if I could get an intelligible sentence out of it all. It was fairly fruitless until I tried one combo, and got the following (obviously incorrect!) translation...
http://img140.imageshack.us/img140/1840/95997474.png
I tried it again, and couldn't duplicate the result.
So, next stop... Vancouver... I'll be sure to bring my trench coat!
-
**3 Solutions to the Liber CCXX II:76 Riddle
**http://img403.imageshack.us/img403/4237/ii76.png
**Solution 1
**
The sequence opens with"4 6 3 8"
4 times 6 is 24
3 times 8 is 24The second row opens with
"x 24 89"
"x" is the 24th letter of the alphabet, and 89 is the 24th prime.
So the key is the English alphabet, in alphabetical order, attached to the prime numbers.
The first 26 primes
1 A 2
2 B 3
3 C 5
4 D 7
5 E 11
6 F 13
7 G 17
8 H 19
9 I 23
10 J 29
11 K 31
12 L 37
13 M 41
14 N 43
15 O 47
16 P 53
17 Q 59
18 R 61
19 S 67
20 T 71
21 U 73
22 V 79
23 W 83
24 X 89
25 Y 97
26 Z 101So, on to deciphering.
The first line is
4638ABK24aLGMOR3YIf we take the first two pairs of numbers as each representing 24 (based on our initial observation that 4 times 6 is 24, and 3 times 8 is 24), then we get
24 24 ABK 24 aLGMOR 3 Y
Using our prime values for the letters, we get
ABK = 2+3+31 = 36
aLGMOR = 2+37+17+41+53+61 = 211
Y = 97Adding it all together, we get:
24+24+36+24+211+3+97 = 419The second line is
x 24 89 RPSTOVAL.If we ignore the “x 24 89” as the cipher, we get
RPSTOVAL = 71+59+47+23+53+13+2+37 = 417
The final answer is then literally and metaphorically “between the lines” – the average between lines 1 and 2 is 418! Abrahadabra!
**Solution 2
**
If we use standard Hebrew gematria on the second row letters, we get:RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 200+80+60+9+70+6+1+30 = 456
On the first row, if we ignore the 4638 and 24 as being part of the embedded cipher clue, we have
ABK, aLGMOR, 3 and Y
Using standard Hebrew gematria on row 2, as we did in Solution 2, we get
23, 344, 3, and 10, which add up to 380
380 + 38 = 418
456 – 38 = 418In other words, once again the average of rows 1 and 2 are 418!
Further, 38 is one of the numbers in the code!
**Solution 3
**
So, what happens if we take the Hebrew alphabet, and line them up to the first 22 primes?2 aleph
3 beth
5 gimel
7 daleth
11 Heh
13 waw
17 zayin
19 cheth
23 theth
29 Yod
31 Kaph
37 Lamed
41 Mem
43 Nun
47 Samek
53 Ayin
59 Peh
61 Tzaddi
67 Qoph
71 Resh
73 Shin
79 TavUsing this cipher, we get the following values for the letters of the first row.
ABK = Aleph-Beth-Kaph = 2+3+31 = 36
ALGMOR = aleph-lamed-gimel-mem-ayin-resh = 2+37+5+41+53+61+29 = 199
3Y = 3 yod = 3 times 29 = 8736+199+87 = 322
That leaves the 4638 and the 24. If we treat the 46 and 38 as ciphers for 24, we get
322+24+24+24 = 394
RPSTOVAL = Resh-Peh-Samekh-Teth-Ayin-Vav-Aleph-Lamed = 71+59+47+23+53+13+2+37 = 305
305 + 23 + 89 = 418!
Now, you’ll notice that that leaves off the “X” of the second row. “X”, the 24th letter, does not have a Hebrew equivalent. But, if we take it to have a value of 24, and we add it’s value to the first row, we get…
418!
So we have one row, the total of all characters being 418 - 24, and another row where all characters total 418 + 24!
@Alrah said
"alrah giggles
Happy Saturnalia Av."
Thanks!
It DOES seem like a nice saturnalia wish to us all!From the team at Babylon translation, or the entity manipulating the signals, to all of us - let's celebrate the kingship of the sun with drinking, misbehaving, and general insane life!