Edwin is a judge and a numerologist. He is married to a woman whose name:
1) Has a "product" that is the same as that for "JUDGE" using the correspondence of letters and numbers below.
A-1 B-2 C-3 D-4 E-5 F-6 G-7 H-8 I-9
J-10 K-11 L-12 M-13 N-14 O-15 P-16 Q-17 R-18
S-19 T-20 U-21 V-22 W-23 X-24 Y-25 Z-26
2) Has no letter in common with "JUDGE" and has no "C" because 3 is his unlucky number.
3) Has its letters in alphabetical order when the first letter and the second letter are interchanged.
What is the English name of the judge´s wife?
I started out by determining the "product" of JUDGE... I assumed that product was meant literally, so I multiplied the values together giving a result of 29400.
I initially started out listing the vowels that were left - A, I, and O, and made the mental note that since A = 1, it wouldn't "count" in the total (thus we could have infinite number of A's if we wished).
I then noted that one of the factors would need to be a factor of 5.
Since E (5) and J (10) were already excluded, that left me with O, T, and Y. I immediately considered the O because it was one of the vowels I knew must be used.
At this point, however, I excluded both O and T because combined we have no way of getting the second factor of 5 we need (the only way to use one is to use the other, and combined leaves us with an impossible numerical value), thus the only letter left was Y.
Factoring this out, you get (5*5)*2*2*2*3*7*7=29400
Since G (7) and U (21) were excluded, the only number left was N (14), of which I needed to have two to take care of both 7s.
That left me with: (5*5)*2*3*(2*7)*(2*7).
Since 2*3 = 6, and we can't logically make an English name with B, C, N, N, and Y no matter how many A's you throw at it (I feel the statement of C being excluded is extraneous), the only solution is FNNY, adding an A you get FANNY which satisfies requirement 3 (AFNNY) (which also may be extraneous, as it is the only logical name to be made).
Solution: FANNY (6*1*14*14*25 = 29400)
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Posted by Lukahn
on 2005-12-09 02:23:20 |