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One of my students pointed out an interesting trick for calculating multiples of $\dfrac{1}{7}$.

Suppose you have memorised $\dfrac{1}{7} = 0.\dot{1}4285\dot{7}$ where the dots mean everything between the dots is repeated.

The important part is the digits after the decimal point: $1,4,2,8,5,7$.

To calculate $\dfrac{2}{7}$, you start with the second lowest digit (which is $2$) and then keep the order of the numbers the same after the 2. So:

$\dfrac{2}{7} = 0.\dot{2}8571\dot{4}$

Similarly to calculate $\dfrac{3}{7}$ you cycle through the digits, starting at the third lowest (which is 4).

Why does this pattern occur?

Jamminermit
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  • Also related: https://math.stackexchange.com/q/4256774/42969 – Martin R Oct 09 '24 at 18:48
  • One answer to this question is that this pattern occurs for fractions with denominator $p$, where $p$ is a prime such that $10$ is a primitive root modulo $p$ – Ben Grossmann Oct 09 '24 at 18:52
  • Next, you can try $1/17$ which is $0588235294117647$ repeating, and multiples of it show the same period with different starting points. Then: in case $1/k$ has period $k-1$, it should also work the same way. – GEdgar Oct 09 '24 at 18:52

1 Answers1

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It can be proved that every rational ultimately ends with trailing $0$ or a cycle.

The reason is simple, there are only $7$ remains in the division by $7$, ie. $0,1,2,3,4,5,6$ so sooner or later one will repeat and you will have a cycle.

  • $1$ divided by $7$ is $0$ remains $1$
  • we offset by $10$ and $10$ divided by $7$ is $1$ remains $3$
  • we offset by $10$ and $30$ divided by $7$ is $4$ remains $2$
  • we offset by $10$ and $20$ divided by $7$ is $2$ remains $6$
  • we offset by $10$ and $60$ divided by $7$ is $8$ remains $4$
  • we offset by $10$ and $40$ divided by $7$ is $5$ remains $5$
  • we offset by $10$ and $50$ divided by $7$ is $7$ remains $1$

and we go to the second line, because the remains $1$ is repeated.

Note that for $\frac 47$ for instance you will simply start the cycle from the line that spells "40 divided by 7", and cycle from there.

Of course this is identical with a larger denominator, just that the cycle may be longer.

By the way, the usual notation is $\frac 17=0.\overline{142857}\cdots\ $ with an overline to mean repeating digits.

zwim
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    I don't see how this answers OP's question. Why does doubling, tripling, etc. simply shift the cycle downstream? Doesn't happen with, say, 1/3. – Randall Oct 09 '24 at 18:45
  • This is because all possible remainders are exhausted in the division by $7$ so multiplying just makes you starts the cycle at a later index. In $\frac 13$ only remainder $1$ is activated. – zwim Oct 09 '24 at 18:50
  • @zwim Ok, but that is the pattern that the question is asking about, which you don't address in your answer – Ben Grossmann Oct 09 '24 at 18:55