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    <p>I can give a hint - repeating decimals in base ten are all fraction with the denominator having at least one prime factors other than two and five. If the denominator contains no prime factors two or five, they can always be represented with a denominator of all nines. Then the nominator is the repeating part and the number of nines is the length of the repeating part.</p> <pre><code>3 _ - = 0.3 9 1 142857 ______ - = ------ = 0.142857 7 999999 </code></pre> <p>If there are prime factors two or five in the denominator, the repeating part starts not at the first position.</p> <pre><code>17 17 ______ -- = ----- = 0.4857142 35 5 * 7 </code></pre> <p>But I cannot remember how to derive the non-repeating part and its length.</p> <p>This seem to translate well to base two. Only fraction with a power of two denominator are non-repeating. This can be easily checked by asserting that only a single bit in the denominator is set.</p> <pre><code>1/2 = 1/10 = 0.1 1/4 = 1/100 = 0.01 3/4 = 11/100 = 0.11 5/8 = 101/1000 = 0.101 </code></pre> <p>All fraction with odd denominators should be repeating and the pattern and its length can be obtained by expressing the fraction with a denominator in the form <code>2^n-1</code>.</p> <pre><code> __ 1/3 = 1/(2^2-1) = 1/11 = 0.01 __ 2/3 = 2/(2^2-1) = 10/11 = 0.10 __ 4/3 =&gt; 1 + 1/3 =&gt; 1.01 __ 10/3 =&gt; 3 + 1/3 =&gt; 11.01 ____ 1/5 = 3/15 = 3/(2^4-1) = 11/1111 = 0.0011 ________ 11/17 = 165/255 = 11/(2^8-1) = 10100101/11111111 = 0.10100101 </code></pre> <p>As for base ten, I cannot tell how to handle denominators containing but not being a power of two - for example 12 = <code>3 * 2^2</code>.</p>
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    1. POAlgorithm for detecting repeating decimals?
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    1. USDaniel Brückner
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    1. CO+1 By this logic then, in base 2, repeating decimals are fractions with denominators having prime factors other than 2 (I knew this). I didn't know that if they had a prime factor 1 it started somewhere other than the first position (that's helpful information!).
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