Divide, add, multiply and understand.

Going through some more old stuff, and thought perhaps I should explain in more detail why I tampered with the text of the Chemical Wedding embedded in A. E. Waite's Real History of the Rosicrucians. One of the characters in this strange work is a "most beautiful Virgin" who supervises the narrator and other guests in the castle where the main action takes place. Towards the end of the narrative of the "Third Day" (Waite, p. 142), after the discussion of a series of moral dilemmas, the narrator asks her name and is met with a numerical riddle:

My name contains six and fifty, and yet hath only eight letters; the third is the third part of the fifth, which added to the sixth will produce a number, whose root shall exceed the third itself by just the first, and it is the half of the fourth. Now the fifth and seventh are equal, the last and first also equal, and make with the second as much as the sixth hath, which contains four more than the third tripled.

As originally printed by Waite, and apparently in the first publication (1690) of the English translation, this read "contains five and fifty" (I have not had a chance to consult the German original). The amendment had nothing to do with the line in Liber Al vel Legis, and everything to do with the solution to the riddle presented by Waite in his 1924 Brotherhood of the Rosy Cross, (p. 168 note) which makes her name ALCHIMIA, taking the number of each letter as its ordinal position in the English or German alphabet. While Waite's reasoning depends on her response to a later question of the narrator, that the seventh (and thus also the fifth) letter "contains . . . as many as there are lords here," the riddle is soluble without this information, thus: Call the letters of her name a, b, c, d, e, f, g, h. These sum to 56. Since "the fifth and seventh are equal, the last and first also equal"

2a + b + c + d + 2e+ f = 56

"the third is the third part of the fifth," to 3c = e, so

2a + b + 7c + d + f = 56

"the third . . . added to the sixth, will produce a number whose root shall exceed the third itself by just the first", so,

sqrt(c + f) = c + a

". . . and it [the root of c+f, or c+a] is half the fourth" so d = 2(c + a), so

4a + b + 9c + f = 56

"the sixth . . . containeth four more than the third tripled" so f = 3c + 4, so

4a + b + 12c + 4 = 56, and sqrt(4c + 4) = c + a, which latter can be rewritten as sqrt 4(c + 1) = c + a, or 2 sqrt (c + 1) = c + a

"the last and the first are also equal, and make with the second, as much as the sixth have." 

Slightly ambiguous, could mean a (or h) + b = f, or a + h + b (= 2a+b) = f. 

If the former (call this case i) then a + b = 3c + 4, so 3a + 15c + 8 = 56, so 3a + 15c = 48, a + 5c = 16. 

If the latter (call this case ii), 2a + b = 3c + 4, so 2a + 15c + 8 = 56, 2a + 15c = 48. 

The reference of numbers to letters strongly suggests that a positive integer solution for all the variables is expected.  At this point, c + 1 has to be a perfect square; which could make c 3, 8, 15, &c.  However if c is more than 3 and a third, a will be negative.  So c = 3.  In case i, a + 15 = 16, so a = 1.  In case ii, 2a + 45 = 48, so a = 1.5, suggesting that the case i reading of the constraint was correct. 

So h = 1, e = 9, g = 9, f = 3 x 3 + 4 = 13, b = 12, f = 2(1+3) = 8, giving 1, 12, 3, 8, 9, 13, 9, 1.

By ordinal position in the German or English alphabet (i.e. treating i and j as different letters), ALCHIMIA. 

If we do not make the assumption about positive integers, it is not possible to resolve the ambiguity noted. Assuming case 1, we take the equations

2 sqrt (c + 1) = a + c, and a + 5c = 16

from the latter, a + c = 16 - 4c; so we can substitute in the first, giving

2 sqrt (c + 1) = 16 - 4c, or sqrt (c + 1) = 8 - 2c c + 1 = 4c^2 - 32c + 64 4c^2 - 33c + 63 = 0

which has two real solutions for c, 3 and 5.25.