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PARTIAL CAROUSEL NUMBERS
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The number D = 076923 behaves a bit like a carousel number. It has 6
digits (again we need the leading 0), and half of the multiples
from 1D, 2D, 3D, ..., 12D are cyclic permutations of D.
For example, 3D = 230769, and 4D = 307692 give the same digits in the same
order. However 2D = 153846, and 5D = 384615 do not. (Note 5D is a cyclic
permutation of 2D. Hmmmmm... check out the other multiples)
So this number, D = 076923, is sort of "half" a carousel number. Not all
the beginning multiples produce cyclic permutations, but half do.
What is true about D = 076923 is that all of the cyclic permutations,
769230, 692307, etc are multiples of D. (Check that.)
DEFINITION
So we might define a PARTIAL CAROUSEL NUMBER to be an integer (perhaps
with leading 0's) with the property that all of its cyclic permutations
are multiples of the number.
CONNECTION WITH PERIODS OF DECIMALS
Again there is a connection with periodic decimals. 1/13 = .076923...
repeating with a period of 6 (half the max. potential period of 12).
Looking at the powers of 10 in U13 (which are the same as the
remainders in the long division process to find the decimal for 1/13)
we get 10 = 10 (mod 13)
100 = 9 (mod 13) - subtract 7 13's = 91
10^3 = 90 = 12 (mod 13)
10^4 = 120 = 3 (mod 13)
10^5 = 30 = 4 (mod 13)
10^6 = 40 = 1 (mod 13).
These are exactly the multiples that "work": 10D, 9D, 12D, 3D, 4D,
(and 1D) give the 6 cyclic permutations of the digits of D.
Further, they are in the order of how much cycling happens:
10D = 769230 cycle or "left shift" by 1
9D = 692307 left shift by 2 from D = 076923
12D = 923076 left shift by 3
etc.
So we are on to something here. We can find more interesting numbers
by not insisting that all the multiples 2X, 3X, ...dX be cyclic
permutations, but only that all cyclic permutations be multiples.
A PROBLEM ARISES
Alas, the inclusion of leading 0's develops into a real problem.
For example, 37 is not a partial carousel number because 73 is not
a multiple of 37. But 037 is!
(shift left 1) 370 = 10 times 37
(shift left 2) 703 = 19 times 37
It turns out that because 37 divides 999 (3 nines) we need the leading
zero to have a 3 digit partial carousel number.
THEOREM
(nearly) EVERY NUMBER WILL BE A PARTIAL CAROUSEL NUMBER
Idea of the proof: Nearly every number divides a Nines number. The
length of the Nines number needed is related to the period of 1/n.
Add enough leading 0's to match the length of that Nines number. You
will now have a partial carousel number.
EXAMPLE Let n = 123. Factoring, n = 3 x 41. It turns out 1/3 has
period 1 and 1/41 has period 5, so 1/123 has period 5. We
claim 00123 has the partial carousel property. (We have added
enough leading 0's to bring the length to 5.)
COROLLARY
Maybe partial carousel numbers aren't all that interesting after all.
Oh well, you can't win them all.