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Mathematics

Perform the long division for 113\dfrac{1}{13}. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 213\dfrac{2}{13}? Now compute 313\dfrac{3}{13}, 413\dfrac{4}{13}, etc. What do you notice?

Whole Numbers

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Answer

Performing long division for 113\dfrac{1}{13} :

13)‾ 0.076923…13))1.000000…‾13‾−0‾13)‾)100000…13)‾−0‾13)‾ 0100000…13)‾1−91‾000…13)‾ 0009000…13)‾11−78‾00…13)‾ 0001200…13)‾)1−117‾0…13)‾ 0000030…13)‾1111−26‾…13)‾ 0000004013)‾11111−39‾13)‾ 00000001…\begin{array}{l} \phantom{13\overline{)}\,}0.076923\ldots \ 13\overline{\smash{\big)}\phantom{)}1.000000\ldots} \ \phantom{13\overline{}}\underline{-0} \ \phantom{13\overline{)})}10\phantom{0000\ldots} \ \phantom{13\overline{)}}\underline{-0} \ \phantom{13\overline{)}\,0}100\phantom{000\ldots} \ \phantom{13\overline{)}1}\underline{-91}\phantom{000\ldots} \ \phantom{13\overline{)}\,000}90\phantom{00\ldots} \ \phantom{13\overline{)}11}\underline{-78}\phantom{00\ldots} \ \phantom{13\overline{)}\,000}120\phantom{0\ldots} \ \phantom{13\overline{)})1}\underline{-117}\phantom{0\ldots} \ \phantom{13\overline{)}\,00000}30\phantom{\ldots} \ \phantom{13\overline{)}1111}\underline{-26}\phantom{\ldots} \ \phantom{13\overline{)}\,000000}40 \ \phantom{13\overline{)}11111}\underline{-39} \ \phantom{13\overline{)}\,0000000}1\ldots \end{array}

⇒ 113=0.076923 076923…=0.076923‾\dfrac{1}{13} = 0.076923 \ 076923 \ldots = 0.\overline{076923}

The repeating block is 076923 (6 digits).

Computing other multiples of 113\dfrac{1}{13} :

Long division for 213\dfrac{2}{13}:

13)‾ 0.153846…13))2.000000…‾13‾−0‾13)‾1200000…13,‾−13‾13)‾ 0070000…13)‾1−65‾000…13)‾ 1115000…13)‾11−39‾00…13)‾ 00001100…13)‾111−104‾0…13)‾))1111160…13)‾11111−52‾…13)‾))1111118013)‾ 000000−78‾13)‾)111111112…\begin{array}{l} \phantom{13\overline{)}\,}0.153846\ldots \ 13\overline{\smash{\big)}\phantom{)}2.000000\ldots} \ \phantom{13\overline{}}\underline{-0} \ \phantom{13\overline{)}1}20\phantom{0000\ldots} \ \phantom{13\overline{,}}\underline{-13} \ \phantom{13\overline{)}\,00}70\phantom{000\ldots} \ \phantom{13\overline{)}1}\underline{-65}\phantom{000\ldots} \ \phantom{13\overline{)}\,111}50\phantom{00\ldots} \ \phantom{13\overline{)}11}\underline{-39}\phantom{00\ldots} \ \phantom{13\overline{)}\,0000}110\phantom{0\ldots} \ \phantom{13\overline{)}111}\underline{-104}\phantom{0\ldots} \ \phantom{13\overline{)}))11111}60\phantom{\ldots} \ \phantom{13\overline{)}11111}\underline{-52}\phantom{\ldots} \ \phantom{13\overline{)}))111111}80 \ \phantom{13\overline{)}\,000000}\underline{-78} \ \phantom{13\overline{)})11111111}2\ldots \end{array}

⇒ 213=0.153846‾\dfrac{2}{13} = 0.\overline{153846}

Long division for 313\dfrac{3}{13}:

13)‾ 0.230769…13))3.000000…‾13‾−0‾13)‾)300000…13‾−26‾13)‾)140000…13‾1−39‾13)‾1111000…13)‾)11−0‾00…13)‾ 00001000…13)‾1111−91‾0…13)‾))1111190…13)‾11111−78‾…13)‾))1111112013)‾ 00000−117‾13)‾)111111113…\begin{array}{l} \phantom{13\overline{)}\,}0.230769\ldots \ 13\overline{\smash{\big)}\phantom{)}3.000000\ldots} \ \phantom{13\overline{}}\underline{-0} \ \phantom{13\overline{)})}30\phantom{0000\ldots} \ \phantom{13\overline{}}\underline{-26} \ \phantom{13\overline{)})1}40\phantom{000\ldots} \ \phantom{13\overline{}1}\underline{-39} \ \phantom{13\overline{)}111}10\phantom{00\ldots} \ \phantom{13\overline{)})11}\underline{-0}\phantom{00\ldots} \ \phantom{13\overline{)}\,0000}100\phantom{0\ldots} \ \phantom{13\overline{)}1111}\underline{-91}\phantom{0\ldots} \ \phantom{13\overline{)}))11111}90\phantom{\ldots} \ \phantom{13\overline{)}11111}\underline{-78}\phantom{\ldots} \ \phantom{13\overline{)}))11111}120 \ \phantom{13\overline{)}\,00000}\underline{-117} \ \phantom{13\overline{)})11111111}3\ldots \end{array}

⇒ 313=0.230769‾\dfrac{3}{13} = 0.\overline{230769}

Similarly, we have:

⇒ 413=0.307692‾\dfrac{4}{13} = 0.\overline{307692}

⇒ 513=0.384615‾\dfrac{5}{13} = 0.\overline{384615}

⇒ 613=0.461538‾\dfrac{6}{13} = 0.\overline{461538}

⇒ 713=0.538461‾\dfrac{7}{13} = 0.\overline{538461}

⇒ 813=0.615384‾\dfrac{8}{13} = 0.\overline{615384}

⇒ 913=0.692307‾\dfrac{9}{13} = 0.\overline{692307}

⇒ 1013=0.769230‾\dfrac{10}{13} = 0.\overline{769230}

⇒ 1113=0.846153‾\dfrac{11}{13} = 0.\overline{846153}

⇒ 1213=0.923076‾\dfrac{12}{13} = 0.\overline{923076}

Observations :

1. The decimal expansions of n13\dfrac{n}{13} form two distinct cyclic groups of 6 digits each, unlike 17\dfrac{1}{7} which has one cyclic group.

2. Group 1 : 113,313,413,913,1013,1213\dfrac{1}{13}, \dfrac{3}{13}, \dfrac{4}{13}, \dfrac{9}{13}, \dfrac{10}{13}, \dfrac{12}{13} all share cyclic permutations of "076923".

3. Group 2 : 213,513,613,713,813,1113\dfrac{2}{13}, \dfrac{5}{13}, \dfrac{6}{13}, \dfrac{7}{13}, \dfrac{8}{13}, \dfrac{11}{13} all share cyclic permutations of "153846".

Hence, 113=0.076923‾\dfrac{1}{13} = 0.\overline{076923} shows cyclic behaviour, but unlike 17\dfrac{1}{7}, it produces two different cyclic blocks among its multiples.

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