## New Merology

Reference: Prof Ian Stewart’s Cabinet of Mathematical Curiosities

New Merology

Let him that hath understanding count the number of the beast; for it is the number of a man, and his number is Six hundred threescore and six. Revelation of St. John 13:18

Or may be not. The Oxyrhynchus Papyri — ancient documents found at Oxyrhynchus in Upper Egypt — include a fragment of the Book of Revelation from the third or fourth century which contains the earliest version of some sections. The number that this papyrus assigns to the Beast is 616, not 666. So much for barcodes being symbols of evil. No  matter, for this puzzle is not about the Beast. It is about an idea that its inventor Lee Sallows, called “New metrology”. Let it be clear that his proposal was not serious, except as a mathematical problem.

The traditional methods for assigning numbers to names, known as gematria, sets $A=1$, $B=2$ upto $Z=26$. Then you add up all the numbers corresponding to the letters of the name. But there are lots of different systems of this kind, and lots of alphabets. Sallows suggested a more rational method based on words that denote numbers. For instance, with the numbering just described, the word ONE becomes $15+14+5=34$. However, the number corresponding to ONE surely  ought to be 1. Worse, no English number denotes its numerological total, a property we will call “perfect”.

Sallows wondered what happens if assign a whole number to each letter, so that as many as, possible of the number words ONE, TWO, and so on are perfect. To make the problem interesting, different letters must be given different values. So you get a whole pile of equations like

$O+N+E=1$

$T+W+O=2$

$T+H+R+E+E=3$

In algebraic unknowns O, N, E, T, W, H, R …And, you must solve them in integers, all distinct.

The equation $O+N+E=1$ tells us that some of the numbers have to be negative. Suppose, for example, that $E=1$ and $N=2$. Then, the equation for ONE tells us that $O=-2$, and similar equations with other number-words imply that $I=4$, $T=7$ and $W=-3$. To make THREE perfect, we must assign values to H and R. If $H=3$, then R has to be -9. FOUR involves two more new letters, F and U. If $F=5$, then $U=10$. Now, $F+I+V+E=5$ leads to $I=-5$. Since SIX contains two new letters, we try SEVEN first, which tells us that $S=8$. Then, we can fill in X from SIX, getting $X=-6$. The equation for EIGHT leads to $G=-7$. Now, all the number names from ONE to TEN are perfect.

The only extra letter in ELEVEN and TWELVE is L. Remarkably, L=11 makes them both perfect. But $T+H+I+R+T+E+E+N=7+3+4+(-9)+7+1+1+2$, which is 16, so we get stuck at this point.

In fact, we always get stuck at this point. If THIRTEEN is perfect, then

$THREE+TEN=THIRTEEN$

and we can remove common letters from both sides. This leads to $E=I$, violating the rule that different letters get different values.

However, we can go the other way and try to make ZERO perfect as well as ONE to TWELVE. Using the choices above, $Z+E+R+O=0$ leads to $Z=10$, but that’s the same as U.

Can you find a different assignment of positive or negative whole-number values to letters, so that all words from ZERO to TWELVE are perfect?

More fun later,

Nalin Pithwa

1. Anubhav Clancy Singh

This took some time, but here it is:-
E 3
F 9
G 6
H 1
I -4
L 0
N 5
O -7
R -6
S -1
T 2
U 8
V -3
W 7
X 11
Z 10

2. Posted January 30, 2016 at 1:53 pm | Permalink | Reply

Great !!! Keep it up !!!

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