142857, the Number That Rotates, math that surprises you lesson

Multiply $142857$ by $2$ and you get $285714$ , the same six digits in the same circular order, read from a different starting point. Multiplying by $3$ , $4$ , $5$ and $6$ does the same, and multiplying by $7$ gives $999999$ . Why would multiplication only move the starting point around a circle? This page follows the long division of $1$ by $7$ , shows that its remainders run through a loop of all six nonzero values, and explains how that loop produces the rotations and why many repeating decimals do not behave this way.

Key terms

Cyclic number

Definition A cyclic number is a positive whole number, written with $n$ digits for some $n \ge 2$ (a leading zero is allowed), such that multiplying it by each of $1, 2, \ldots, n$ gives a product whose $n$ digits are a rotation of its own digits. A rotation keeps the digits in the same circular order and starts reading them at another place, as $142857$ read from the $2$ gives $285714$ . The number $142857$ is a cyclic number with six digits, and its seventh multiple is $7 \cdot 142857 = 999999$ , which is not a rotation. The explanation is that $142857$ is the repeating block of $1/7 = 0.142857142857\ldots$ , and for $k = 1, \ldots, 6$ the product $k \cdot 142857$ is the repeating block of $k/7$ . The long division of each $k/7$ runs around the same loop of remainders as the division of $1$ by $7$ , entering it at remainder $k$ , so every block is the same six digits read from a different place.

In plain words Put the six digits of $142857$ on a circle. Multiplying by any of $1$ to $6$ only changes where you start reading, because the products are the repeating blocks of $1/7$ through $6/7$ , and all six blocks come from one loop of long division.

Example The table lists each product and the digit of the circle where its reading starts.

Multiplier	Product	Reading starts at
$1$	$142857$	$1$
$2$	$285714$	$2$
$3$	$428571$	$4$
$4$	$571428$	$5$
$5$	$714285$	$7$
$6$	$857142$	$8$
$7$	$999999$	not a rotation

Beyond $7$ the products have seven digits, for example $8 \cdot 142857 = 1142856$ , so the rotations belong to the multipliers $1$ to $6$ only.

Long division

Definition Long division finds the decimal digits of a fraction $p/q$ , with whole numbers $0 < p < q$ , one digit at a time. The first remainder is $r = p$ . At each step, ten times the current remainder is divided by $q$ , which gives a quotient $D$ and a new remainder $r'$ with

\begin{gathered} 10r = qD + r', \\[0.4em] 0 \le r' < q. \end{gathered}

The quotient $D$ is the next decimal digit, and it is a whole number from $0$ to $9$ because $10r < 10q$ . The new remainder $r'$ starts the next step. If a remainder is $0$ , every later digit is $0$ and the decimal ends.

In plain words Each step asks how many times $q$ fits into ten times what is left over. The answer is the next digit, and whatever is left over again is carried into the next step.

Example For $1/7$ the six steps are

Remainder	Ten times it	Digit	New remainder
$1$	$10 = 7 \cdot 1 + 3$	$1$	$3$
$3$	$30 = 7 \cdot 4 + 2$	$4$	$2$
$2$	$20 = 7 \cdot 2 + 6$	$2$	$6$
$6$	$60 = 7 \cdot 8 + 4$	$8$	$4$
$4$	$40 = 7 \cdot 5 + 5$	$5$	$5$
$5$	$50 = 7 \cdot 7 + 1$	$7$	$1$

The digits come out in the order $1, 4, 2, 8, 5, 7$ , and the last new remainder is $1$ , the remainder the division started with. So the same six steps happen again and again, and

\frac{1}{7} = 0.142857\,142857\ldots

Remainder cycle

Definition In the long division of $p$ by $q$ , with $0 < p < q$ , each remainder is one of the whole numbers $0, 1, \ldots, q - 1$ , and it alone decides the next digit and the next remainder. There are only $q$ possible remainders, so among the first $q + 1$ remainders some value appears twice, and from its first appearance on, the same remainders and digits repeat forever in the same order. That repeating loop of remainders is the remainder cycle. When $q$ has no factor $2$ or $5$ , two different remainders never lead to the same next remainder and no remainder is ever $0$ , so the loop closes at the starting remainder $p$ and the decimal repeats from its first digit. Two fractions with denominator $q$ whose starting remainders lie on the same cycle have repeating blocks that are rotations of each other.

In plain words The remainder is the entire memory of a long division. When an old remainder comes back, the division has no choice but to repeat everything that followed it the first time.

Example For $1/7$ the remainders run

\begin{aligned} &1 \to 3 \to 2 \to 6 \\[0.4em] &\quad \to 4 \to 5 \to 1, \end{aligned}

a loop through all six nonzero remainders below $7$ . The division of $3$ by $7$ starts at remainder $3$ , which is already on this loop, so it produces the digits that followed remainder $3$ in the division of $1$ by $7$ ,

\frac{3}{7} = 0.428571\,428571\ldots

That block is $3 \cdot 142857 = 428571$ . Every $k/7$ with $k$ from $1$ to $6$ starts somewhere on the same loop, which is why all six blocks are rotations of $142857$ .

Period of a repeating decimal

Definition A decimal repeats if, from some digit on, one block of digits repeats forever, and its period is the number of digits in the shortest such block. A bar over the block marks it, as in $1/7 = 0.\overline{142857}$ , which has period $6$ . The decimal of every fraction of whole numbers either ends or repeats. Let $p/q$ be a fraction in lowest terms with $0 < p < q$ , where $q$ has no factor $2$ or $5$ . Then the decimal repeats from the first digit after the point, and its period is the smallest whole number $k \ge 1$ for which $q$ divides $10^k - 1$ . With the block written as a $k$ -digit whole number $B$ , leading zeros included,

\frac{p}{q} = \frac{B}{10^k - 1}.

The period is at most $q - 1$ , because the loop holds only nonzero remainders, and it equals $q - 1$ only when $q$ is a prime.

In plain words The period counts how many digits pass before the decimal starts over. Multiplying by $10^k$ moves one whole block in front of the decimal point, and subtracting the original number removes the endless tail.

Example Let $x = 0.\overline{142857}$ . Moving the point six places gives

\begin{aligned} 10^6 x &= 142857.\overline{142857}, \\[0.4em] 10^6 x - x &= 142857, \end{aligned}

so $999999x = 142857$ and $x = 142857/999999 = 1/7$ . None of $9$ , $99$ , $999$ , $9999$ and $99999$ is divisible by $7$ , while $999999 = 7 \cdot 142857$ , so the period of $1/7$ is $6 = 7 - 1$ , the largest possible for the denominator $7$ . Many denominators fall short of the largest possible period. The decimal $1/3 = 0.\overline{3}$ has period $1$ , the decimal $1/11 = 0.\overline{09}$ has period $2$ , and the decimal $1/13 = 0.\overline{076923}$ has period $6$ , half of the largest possible $12$ .

Check yourself

Use the remainder loop of $1/7$ to write $5/7$ as a decimal without dividing again, and compare the result with $5 \cdot 142857$ .

Show the answer

The division of $5$ by $7$ starts at remainder $5$ . In the division of $1$ by $7$ , remainder $5$ produced the digit $7$ , since $50 = 7 \cdot 7 + 1$ , and led back to remainder $1$ , after which the digits were $1, 4, 2, 8, 5$ . So

\frac{5}{7} = 0.\overline{714285},

and indeed $5 \cdot 142857 = 714285$ , the six digits of $142857$ read from the $7$ .

Change the denominator to $13$ . The long division gives $1/13 = 0.\overline{076923}$ . Is $076923$ a cyclic number?

Show the answer

No. The remainders of $1/13$ run $1, 10, 9, 12, 3, 4$ and then return to $1$ , so this loop holds only six of the twelve nonzero remainders. The product $2 \cdot 076923 = 153846$ is not a rotation of $076923$ , so the number fails already at the multiplier $2$ . The division of $2$ by $13$ starts at remainder $2$ , which lies on a second loop $2, 7, 5, 11, 6, 8$ , and it gives $2/13 = 0.\overline{153846}$ .

All twelve multipliers follow the same pattern. Multiplying $076923$ by $1$ , $3$ , $4$ , $9$ , $10$ or $12$ , the remainders on the first loop, gives rotations of $076923$ , while multiplying by $2$ , $5$ , $6$ , $7$ , $8$ or $11$ gives rotations of $153846$ , and $13 \cdot 076923 = 999999$ . In general, for a denominator $q$ with no factor $2$ or $5$ , a multiple $m \cdot B$ of the block $B$ of $1/q$ , with $m$ from $1$ to $q - 1$ , is a rotation of $B$ exactly when the remainder $m$ lies on the loop of $1$ . Every multiplier from $1$ to $q - 1$ gives a rotation only when that loop contains every nonzero remainder, which happens for $7$ but not for $13$ .

Extend the pattern to $17$ . The long division of $1$ by $17$ has period $16$ and the block $0588235294117647$ . What should happen when this block is multiplied by $2$ to $16$ , and by $17$ ?

Show the answer

A period of $16$ means that the remainder loop of $1/17$ holds $16$ different remainders. The only nonzero remainders below $17$ are $1$ to $16$ , so the loop contains all of them. Each $k/17$ with $k$ from $1$ to $16$ therefore enters the same loop, and its block is a rotation of $0588235294117647$ .

Write $B$ for this block. Then $k/17 = kB/(10^{16} - 1)$ , and $kB$ is less than $10^{16} - 1$ for $k \le 16$ , so $kB$ , written with $16$ digits, is exactly the block of $k/17$ . Every product from $2B$ to $16B$ is therefore a rotation of $B$ . For example, $2B = 1176470588235294$ , which is $B$ read from its first $1$ . Multiplying by $17$ gives

17B = 10^{16} - 1,

which is written as sixteen nines. So $0588235294117647$ is a cyclic number with $16$ digits, as long as its leading zero is kept. The same reasoning works for every prime $q$ for which $1/q$ has period $q - 1$ , such as $7$ , $17$ , $19$ and $23$ .