How to Calculate Decimal Places Before Repeating Digits in a Fraction

decimal-expansionfractions

If you have a fraction such as $$\frac{7}{26}=0.269230\overline{769230}$$ where there are a number of digits prior to the repeating section, how can you tell how many digits there will be given just the fraction?

I believe I could run through the standard long division algorithm until I come across the same remainder for the second time and then use the location of the first instance of this remainder to calculate the number of digits before the repeating section, but this feels very cumbersome.

After lots of reading online, I came across what looks like a formula for it from Wolfram MathWorld:

When a rational number $\frac{m}{n}$ with $(m,n)=1$ is expanded, the period begins after ${s}$ terms and has length ${t}$, where ${s}$ and ${t}$ are the smallest numbers satisfying $10^s\equiv10^{s+t}\pmod{n}$.

I know how to calculate the length of the period of a fraction, and so in the case of my original fraction we have $10^s\equiv10^{s+6}\pmod{26}$, but I don't know how to solve for ${s}$ in this equation!

So there are really two questions here – the one in the title, and a sneaky one about how to take logs in a modulo arithmetic equation.

Best Answer

Rewrite the fraction as $$\frac{m}{n}=\frac{p}{10^sq}$$ where $p,q$ are coprime and $q$ is not divisible by $2$ or $5$ while $p$ is not divisible by $10$. Computing $s$ (the pre-period) is easy; it is the larger of the number of times $2$ divides $n$ and the number of times $5$ divides $n$. Then we want the smallest $t$ such that $10^t\equiv 1\;(\bmod\;q)$. By Fermat's little theorem, we have $10^{\varphi(q)}\equiv 1\;(\bmod\;q)$, thus $\;t|\varphi(q)$ so it suffices to check the divisors of $\varphi(q)$.

Related Solutions

[Math] How do we find a fraction with whose decimal expansion has a given repeating pattern

Suppose you want to have a number $x$ whose decimal expansion is $0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$. That is it has a period of length $k$, with digits $a_1$, $a_2,\ldots,a_k$.

Let $n = a_1a_2\cdots a_k$ be the integer given by the digits of the period. Then $$\begin{align*} \frac{n}{10^{k}} &= 0.a_1a_2\cdots a_k\\ \frac{n}{10^{2k}} &= 0.\underbrace{0\cdots0}_{k\text{ zeros}}a_1a_2\cdots a_k\\ \frac{n}{10^{3k}} &= 0.\underbrace{0\cdots0}_{2k\text{ zeros}}a_1a_2\cdots a_k\\ &\vdots \end{align*}$$ So the number you want is $$\sum_{r=1}^{\infty}\frac{n}{10^{rk}} = n\sum_{r=1}^{\infty}\frac{1}{(10^k)^r} = n\left(\frac{\quad\frac{1}{10^k}\quad}{1 - \frac{1}{10^k}}\right) = n\left(\frac{10^k}{10^k(10^k - 1)}\right) = \frac{n}{10^k-1}.$$ Since $10^k$ is a $1$ followed by $k$ zeros, then $10^k-1$ is $k$ 9s. So the fraction with the decimal expansion $$0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$$ is none other than $$\frac{a_1a_2\cdots a_k}{99\cdots 9}.$$

Thus, $0.575757\cdots$ is given by $\frac{57}{99}$. $0.837168371683716\cdots$ is given by $\frac{83716}{99999}$, etc.

If you have some decimals before the repetition begins, e.g., $x=2.385858585\cdots$, then first multiply by a suitable power of $10$, in this case $10x = 23.858585\cdots = 23 + 0.858585\cdots$, so $10x = 23 + \frac{85}{99}$, hence $ x= \frac{23}{10}+\frac{85}{990}$, and simple fraction addition gives you the fraction you want.

And, yes, there is always a solution and it is always a rational.

[Math] How to evalulate a fraction (example 1/998001) accurately to 100 decimal places and then display it in maxima/wxmaxima?

$$ \frac{1}{(1000-1)^2} =\frac{10^{-6}}{(1-\frac{1}{1000})^2} =\frac{10^{-6}}{(1-x)^2} =10^{-6}\left(\sum_{n=0}^{\infty}x^n\right)^2 =10^{-6}\sum_{n=1}^{\infty}nx^{n-1} $$ for $x=.001$, which has no carries in its decimal expansion for $n<1000$. Thus the first $1000$ triplets ($3000$ digits after the decimal place) will be $0.000\;001\;002\;003\;\dots\;996\;997\;999\;000$, which brings us up to the first digit which receives a carry. If you want to see it without the scientific notation, add one before formatting. In sage:

(1+(10^3-1)^(-2)).n(digits=3001) # ignore the leading one in the output!

or in scientific notation:

((10^3-1)^(-2)).n(digits=3001) # note the e-6 at the end, meaning times 10^{-6}

As to the formula above, it can be derived by differentiation: $$ \left(\sum_{n=0}^{\infty}x^n\right)^2 =(1-x)^{-2} =\frac{d}{dx}(1-x)^{-1} =\frac{d}{dx}\left(\sum_{n=0}^{\infty}x^n\right) =\sum_{n=1}^{\infty}nx^{n-1} \qquad\text{for} \qquad|x|<1 $$

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