142857 - A remarkable number

Surprising properties of cyclic numbers.
number theory

June 17, 2020

Cyclic Numbers

As is well known—but perhaps not as well as it should be!—the number \(142,857\) has a remarkable property:

\[ 142857×1=142857 \\ 142857×2=285714 \\ 142857×3=428571 \\ 142857×4=571428 \\ 142857×5=714285 \\ 142857×6=857142 \\ \]

In other words,the first six multiples of \(142,857\) are obtained simply by cyclically permuting its digits. The number \(142,857\) is thus known as a cyclic number, that is,an \(n\)-digit number(possibly having some initial digits \(0\)) whose first \(n\) multiples are given by cyclically permuting the original number in all possible ways. The existence of such a remarkable number immediately raises further questions.

First,do there exist other cyclic numbers?

The answer is yes:

\[ 0,588,235, 294,117,647 \]

is the next smallest cyclic number,having \(16\) digits; multiplying it by any number from \(1\) to \(16\) simply cycles its digits!

The next question that arises, then, is: Are there infinitely many cyclic numbers?
This, remarkably, is an unsolved problem.

It turns out that every cyclic number in base \(10\) is given by the repeating pattern in the decimal expansion of \(1/p\) for a prime \(p\) such that \(10\) is a primitive root modulo \(p\) (i.e., the first \(p−1\) powers of ten—\(10^0, 10^1, \dots,10^{p−2}\)—have distinct remainders when divided by \(p\)). For example, that \(142,857\) is a cyclic number is related to the fact that \(1,10,10^2,10^3,10^4\), and \(10^5\) all yield distinct remainders when divided by \(7\),namely, \(1,3,2,6,4\), and \(5\),respectively.

Exactly nine primes smaller than \(100\) generate cyclic numbers: \(7, 17, 19, 23, 29,47, 59. 61, 97\).

Other surprising properties

  1. When a cyclic number is multiplied by its generating prime, the product is always a row of \(9\)’s. For instance, \(142,857\) times \(7\) is \(999,999\). This provides another way to search for cyclics: divide a prime, \(p\), into a row of \(9\)’s until there is no remainder. If the quotient has \(p - 1\) digits, it is a cyclic number.

  2. Even less expected is the fact that every cyclic (or any of its cyclic permutations), when split in half, gives two numbers that add to a row of \(9\)’s. For example, \(142 + 857 = 999\). For another example, split the cyclic generated by \(1/17\) into halves and add: \(05882352 + 94117647 = 9999999\).


Martin Gardner. Cyclic numbers. In Mathematical Circus: More Puzzles, Games, Paradoxes and Other Mathematical Entertainments from Scientific American. Vintage Books, New York, 1981.

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