Midy's theorem

From Wikipedia, the free encyclopedia
Jump to: navigation, search

In mathematics, Midy's theorem, named after French mathematician E. Midy,[1][2] is a statement about the decimal expansion of fractions a/p where p is a prime and a/p has a repeating decimal expansion with an even period (sequence A028416 in OEIS). If the period of the decimal representation of a/p is 2n, so that

\frac{a}{p}=0.\overline{a_1a_2a_3\dots a_na_{n+1}\dots a_{2n}}

then the digits in the second half of the repeating decimal period are the 9s complement of the corresponding digits in its first half. In other words,

a_i+a_{i+n}=9 \,
a_1\dots a_n+a_{n+1}\dots a_{2n}=10^n-1. \,

For example,

\frac{1}{13}=0.\overline{076923}\text{ and }076+923=999. \,
\frac{1}{17}=0.\overline{0588235294117647}\text{ and }05882352+94117647=99999999. \,

Extended Midy's theorem[edit]

If k is any divisor of the period of the decimal expansion of a/p (where p is again a prime), then Midy's theorem can be generalised as follows. The extended Midy's theorem[3] states that if the repeating portion of the decimal expansion of a/p is divided into k-digit numbers, then their sum is a multiple of 10k − 1.

For example,

\frac{1}{19}=0.\overline{052631578947368421} \,

has a period of 18. Dividing the repeating portion into 6-digit numbers and summing them gives


Similarly, dividing the repeating portion into 3-digit numbers and summing them gives


Midy's theorem in other bases[edit]

Midy's theorem and its extension do not depend on special properties of the decimal expansion, but work equally well in any base b, provided we replace 10k − 1 with bk − 1 and carry out addition in base b.

For example, in octal

& \frac{1}{19}=0.\overline{032745}_8 \\[8pt]
& 032_8+745_8=777_8 \\[8pt]
& 03_8+27_8+45_8=77_8.

In duodecimal (using inverted two and three for ten and eleven, respectively)

& \frac{1}{19}=0.\overline{076\mathcal{E}45}_{12} \\[8pt]
& 076_{12}+\mathcal{E}45_{12}=\mathcal{EEE}_{12} \\[8pt]
& 07_{12}+6\mathcal{E}_{12}+45_{12}=\mathcal{EE}_{12}

Proof of Midy's theorem[edit]

Short proofs of Midy's theorem can be given using results from group theory. However, it is also possible to prove Midy's theorem using elementary algebra and modular arithmetic:

Let p be a prime and a/p be a fraction between 0 and 1. Suppose the expansion of a/p in base b has a period of , so

& \frac{a}{p} = [0.\overline{a_1a_2\dots a_\ell}]_b \\[6pt]
& \Rightarrow\frac{a}{p}b^\ell = [a_1a_2\dots a_\ell.\overline{a_1a_2\dots a_\ell}]_b \\[6pt]
& \Rightarrow\frac{a}{p}b^\ell = N+[0.\overline{a_1a_2\dots a_\ell}]_b=N+\frac{a}{p} \\[6pt]
& \Rightarrow\frac{a}{p} = \frac{N}{b^\ell-1}

where N is the integer whose expansion in base b is the string a1a2...a.

Note that b  − 1 is a multiple of p because (b  − 1)a/p is an integer. Also bn−1 is not a multiple of p for any value of n less than , because otherwise the repeating period of a/p in base b would be less than .

Now suppose that  = hk. Then b  − 1 is a multiple of bk − 1. (To see this, substitute x for bk; then b = xh and x − 1 is a factor of xh − 1. ) Say b  − 1 = m(bk − 1), so


But b  − 1 is a multiple of p; bk − 1 is not a multiple of p (because k is less than  ); and p is a prime; so m must be a multiple of p and


is an integer. In other words,

N\equiv0\pmod{b^k-1}. \,

Now split the string a1a2...a into h equal parts of length k, and let these represent the integers N0...Nh − 1 in base b, so that

N_{h-1} & = [a_1\dots a_k]_b \\
N_{h-2} & = [a_{k+1}\dots a_{2k}]_b \\
& {}\  \   \vdots \\
N_0 & = [a_{l-k+1}\dots a_l]_b

To prove Midy's extended theorem in base b we must show that the sum of the h integers Ni is a multiple of bk − 1.

Since bk is congruent to 1 modulo bk − 1, any power of bk will also be congruent to 1 modulo bk − 1. So

\Rightarrow N \equiv \sum_{i=0}^{h-1}N_i \pmod{b^k-1}
\Rightarrow \sum_{i=0}^{h-1}N_i \equiv 0 \pmod{b^k-1}

which proves Midy's extended theorem in base b.

To prove the original Midy's theorem, take the special case where h = 2. Note that N0 and N1 are both represented by strings of k digits in base b so both satisfy

0 \leq N_i \leq b^k-1. \,

N0 and N1 cannot both equal 0 (otherwise a/p = 0) and cannot both equal bk − 1 (otherwise a/p = 1), so

0 < N_0+N_1 < 2(b^k-1) \,

and since N0 + N1 is a multiple of bk − 1, it follows that

N_0+N_1 = b^k-1. \,


From the above,

\frac{am}{p} is an integer

Thus m = 0 \pmod{p}

And thus for k = \frac{l}{2}

b^{\frac{l}{2}}+ 1 = 0 \pmod{p}

For k = \frac{l}{3} and is an integer

b^{\frac{2}{3}l} + b^{\frac{l}{3}} + 1 = 0 \pmod{p}

and so on.


  1. ^ Leavitt, William G. (June 1967). "A Theorem on Repeating Decimals". The American Mathematical Monthly (Mathematical Association of America) 74 (6): 669–673. doi:10.2307/2314251. 
  2. ^ Kemeny, John. "The Secret Theorem of M. E. Midy = Casting In Nines". Retrieved 27 November 2011. 
  3. ^ Bassam Abdul-Baki, Extended Midy's Theorem, 2005.


  • Rademacher, H. and Toeplitz, O. The Enjoyment of Mathematics: Selections from Mathematics for the Amateur. Princeton, NJ: Princeton University Press, pp. 158–160, 1957.
  • E. Midy, "De Quelques Propriétés des Nombres et des Fractions Décimales Périodiques". College of Nantes, France: 1836.
  • Ross, Kenneth A. "Repeating decimals: a period piece". Math. Mag. 83 (2010), no. 1, 33–45.

External links[edit]