Every letter of the alphabet is given a nonzero numerical value which need not be an integer. The value of a word is the product of the values of its letters. For example, if c=2, a=4, and t=3, then "cat" has the value 2*4*3=24.

Suppose that every pair of homophones has equal values (thus, ate=eight, implying a=igh).

Show that every letter of the alphabet has value 1.

Actually, fourth thoughts from the looks of it, but I missed the original posting of this puzzle. Still, I want to post my ideas ..

Finding a pair of homophones where one contains all the letters of the other but one, as in ROTE/WROTE, means of course that the added letter must be equal to one. I did as many letters as I could in this fashion:

ISLE/AISLE (A=1)
JAM/JAMB (B=1)
SENT/SCENT (C=1)
SLIGHT/SLEIGHT (E=1)
REIN/REIGN (G=1)
CORAL/CHORAL (H=1)
HOSTEL/HOSTILE (I=1)
SIC/SICK (K=1)
HAVE/HALVE (L=1)
DAM/DAMN (N=1)
TO/TOO (O=1)
PINCHER/PINSCHER (S=1)
GAGE/GAUGE (U=1)
HORDE/WHORED (W=1)
SPADE/SPAYED (Y=1)

A B C _ E _ G H I _ K L _ N O _ _ _ S _ U _ W _ Y _

Then, having a group of letters shown to be equal to one makes it easy to find a few more:

DAMNED/DAMMED (M=N=1)
BEAUT/BUTTE (T=A=1)
LACKS/LAX (X=CK=1)
PRIES/PRIZE (Z=S=1)

A B C _ E _ G H I _ K L M N O _ _ _ S T U _ W X Y Z

Other letters I couldn't find a "pretty" example that got the letter all by itself with a product equal to one on the other side, but factoring out extra terms was still relatively simple:

MEDAL/MEDDLE (A=DE; A=E=1 .: D=1)
GENE/JEAN (GE=JA; A=E=G=1 .: J=1)
CUE/QUEUE (C=QUE; C=E=U=1 .: Q=1)
BARD/BARRED (RE=1; E=1 .: R=1)

A B C D E _ G H I J K L M N O _ Q R S T U _ W X Y Z

I had the most trouble with F, P, and V. It was easy to find numerous examples showing F=P (PROPHET/PROFIT, etc), but solving for one without the other was a little trickier.

Finally, I came up with:

OFFAL/AWFUL (OF=WU; O=U=W=1 .: F=1)
PROPHET/PROFIT (PHE=FI; E=F=I=H=1 .: P=1)

A B C D E F G H I J K L M N O P Q R S T U _ W X Y Z

That just leaves V. Actually, I didn't really find an example for V yet, but I'm still looking. Obviously, there are many ways to prove most of these letters, but I chose the simplest, one-letter-difference pairs first, and of those the most interesting ones (rather than just NOT/KNOT and RITE/WRITE for all of them) I could think of. Fun problem.

Posted by DJ on 2005-01-19 19:13:53