# Re: Powers that Be (was Year of Destiny?!)

Tue, 21 Sep 1999 14:33:30 -0600 (MDT)

Vernon,

During my childhood I was fascinated by the reports of Ivan Panin's work
on Bible numerics, in which he had found so many features of numerical
values of words adding up to multiples of seven. Later I became suspicious
of the significance claimed for these observations when I wondered why
they were not more widely known and when I noted that different types of
features were claimed for different passages, which is not remarkable
since any such computation should have a one-in-seven chance of yielding a
multiple of seven. Also the proponents used these methods to claim support
for the readings of the Textus Receptus, even where there was overwhelming
evidence of later insertions into the original text. Thus I am too
cautious to use numerology to support the inspiration of Scripture unless
I am convinced that no skeptic can come up with another credible
explanation. I consider a numerological argument to be very weak if it is
based on features that can only be found in one verse.

You devoted a lot of space in a recent posting to what is not a very
remarkable feature at all. Your tables of differences between word values
were totally unnecessary. As soon as it is observed that six of the seven
words in Gen. 1:1 have numerical values that are congruent to two modulo
three, it immediately follows that their differences can be written as
combinations of 105 and 99 since any multiple of three can be expressed in
this form (since 3 is the greatest common divisor of 105 and 99). The
probability that a randomly chosen set of seven integers will contain a
set of at least six that are congruent to each other modulo three is
slightly greater than two percent. Thus such a feature would not be
extremely rare. Furthermore it is even more likely in a sentence with two
direct objects. (In Gen. 1:1 each object is preceded by the object marker
'eth, in one instance with the value-6 prefix waw (and).) In the one
instance where a word's value was not congruent to two modulo three, you
replaced it by a number that was, thus enabling you to complete your
table. Since 913 leaves a remainder of 1 when divided by 3, you simply
subtracted a number (500) that was congruent to two modulo three to give
you a number congruent to two modulo three, but you could have
accomplished the same thing by using any number other than 500 with this
property.

Gordon Brown
Department of Mathematics