   SEARCH HOME Math Central Quandaries & Queries  Question from Vir: Years ago I (re?)discovered 'cyclic division'. For example: if you arrange the number along a circle and put the number 142857 at the centre all the numbers taken cyclically, starting with 1, are fully divisible by 37. Whatever the starting point of this number, it remains fully divisible by 37. what is more, the number can be formed by taking the digits clockwise as well as anti-clockwise..This I call "full cyclic divisibility". In many cases, only clockwise cyclic divisibility is possible. But I have not come across a case where ONLY anti-clockwise divisibility occurs. Thus clockwise cyclic divisibility seems to be favoured. Could this be construed as a sign of chirality in mathematics?.. Your cyclic division is a lovely observation! Because deep philosophical questions are not my speciality, I’ll not comment on your chirality question except to observe that in Arabic texts, numbers are written in an order (with the units place on the left) that is the reverse of the order we use (with the units place on the right). Perhaps incompatible chirality explains today’s religious problems.

The cyclic division you observe is an immediate consequence of our decimal number system. The whole story can be found on the web under the heading “repeated decimals.” (You should also check out the interesting side topic “Midy’s theorem”) I’ve attached a worksheet that I discussed at a math workshop for students in grades 7 to 12. The worksheet was based on a more detailed discussion by Kenneth A. Ross: “Repeating decimals: a period piece". Mathematics Magazine 83:1 (Feb. 2010), 33–45. Here is a very brief version of it using your number 142857.

Note that
1/7 = .142857172857…
with the block of period 6 (namely 142657) repeating for ever.
Multiples of 1/7 are obtained by cyclically permuting the digits in the block:
2/7 = .285714…
3/7 = .428571…
and so on.
Thus your number is just (10^6)*(1/7) - (1/7) = 999,999 * (1/7).

Of course 999,999 is divisible by 37 (and also by 27, 7, 11, and 13), and the same will hold with n/7 replacing 1/7 (which cyclically permutes the digits of your number).
The fact that 758241 is also divisible by 37 (and by 27 and 11, but not by 13) seems to be simply a coincidence; the reason why each permutation of its digits is still divisible by 37 is similar but a bit more complicated because the repeating decimal is not in the form 1/p where p is a prime number. Here one works with n/91; for example .758241… = (69)/(91), while .582417 = 53/91, .824175 = 75/91, …, and .175824 = 16/91. For that part of the explanation one needs to understand how this theory is related to mod 10 arithmetic. The attached worksheet should be enough to get you started.

Chris

Vir wrote back

Dear Chris
Many thanks for the reply Some doubts/ queries remain. Especially: The
proposition that Anticlockwise Divisibility never occurs by itself; (ie; it
is always accompanied by Clockwise Divisibility) is verifiable. Is this
proposition correct? Of course, the question relates only to our decimal
numbering system.

Best regards
Vir

As far as I can tell, clockwise and anticlockwise are not well defined — write your number clockwise on a transparent page and turn the page over; suddenly your number is anticlockwise! Let’s look at an example: Cyclic permutations of the number 648351 have 27 as their common divisor (which is almost meaningless because ALL permutations of these six digits are divisible by 9 because they sum to a multiple of 9).

But cyclic permutations of the reverse number 153846 have 76923 (= 27*7*407) as their common divisor. (That 6-digit number comes from multiples of 2/13.)

Chris      Math Central is supported by the University of Regina and the Imperial Oil Foundation.