For any integer base B
greater than 2, the multiple
s of the
-1, when represented in base B
will always have
s be a multiple of B
-1. If B
-1 is a square number
(4,9,16,...) then the square root
-1 will also have this
What you say? Huh?
Most everyone knows of the 'if the number is divisible by 9, the
digits will add up to nine' rule. 81 (9 * 9) is has this property,
as does 47115 (9 * 5235). It may take multiple steps to get to the
final 9, but its always there:
4 + 7 + 1 + 1 + 5 = 18
1 + 8 = 9
9 also happens to be a square number (32 = 9),
and thus 3 has this property too (see
How to determine whether a number is divisible by 3 for a proof).
While this is all well and good, the question of in base 5, does this
work with the number 4? The number 4810
is represented as 1215, and 1 + 2 + 1 = 4.
This appears to hold true for all bases.
The primary 'use' for this is in the optimization of the search
space in self-describing numbers (such as 2100010006) in which
the number has an equal number of digits as the base it is in, and
the sum of the digits is equal to the base. Thus, the number must
be 1 + n(B-1).