If k is any divisor of h (where h is the number of digits 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^[2] 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 10^k − 1.

For example,

${\displaystyle {\frac {1}{19}}=0.{\overline {052631578947368421}}}$ 

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

${\displaystyle 052631+578947+368421=999999.}$ 

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

${\displaystyle 052+631+578+947+368+421=2997=3\times 999.}$ 

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 10^k − 1 with b^k − 1 and carry out addition in base b.

For example, in octal

${\displaystyle {\begin{aligned}&{\frac {1}{19}}=0.{\overline {032745}}_{8}\\[8pt]&032_{8}+745_{8}=777_{8}\\[8pt]&03_{8}+27_{8}+45_{8}=77_{8}.\end{aligned}}}$ 

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

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

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 ℓ, so

${\displaystyle {\begin{aligned}&{\frac {a}{p}}=[0.{\overline {a_{1}a_{2}\dots a_{\ell }}}]_{b}\\[6pt]&\Rightarrow {\frac {a}{p}}b^{\ell }=[a_{1}a_{2}\dots a_{\ell }.{\overline {a_{1}a_{2}\dots a_{\ell }}}]_{b}\\[6pt]&\Rightarrow {\frac {a}{p}}b^{\ell }=N+[0.{\overline {a_{1}a_{2}\dots a_{\ell }}}]_{b}=N+{\frac {a}{p}}\\[6pt]&\Rightarrow {\frac {a}{p}}={\frac {N}{b^{\ell }-1}}\end{aligned}}}$ 

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

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

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

${\displaystyle {\frac {a}{p}}={\frac {N}{m(b^{k}-1)}}.}$ 

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

${\displaystyle {\frac {am}{p}}={\frac {N}{b^{k}-1}}}$ 

is an integer. In other words,

${\displaystyle N\equiv 0{\pmod {b^{k}-1}}.}$ 

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

${\displaystyle {\begin{aligned}N_{h-1}&=[a_{1}\dots a_{k}]_{b}\\N_{h-2}&=[a_{k+1}\dots a_{2k}]_{b}\\&{}\ \ \vdots \\N_{0}&=[a_{l-k+1}\dots a_{l}]_{b}\end{aligned}}}$ 

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

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

${\displaystyle N=\sum _{i=0}^{h-1}N_{i}b^{ik}=\sum _{i=0}^{h-1}N_{i}(b^{k})^{i}}$ 

${\displaystyle \Rightarrow N\equiv \sum _{i=0}^{h-1}N_{i}{\pmod {b^{k}-1}}}$ 

${\displaystyle \Rightarrow \sum _{i=0}^{h-1}N_{i}\equiv 0{\pmod {b^{k}-1}}}$ 

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

${\displaystyle 0\leq N_{i}\leq b^{k}-1.}$ 

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

${\displaystyle 0<N_{0}+N_{1}<2(b^{k}-1)}$ 

and since N₀ + N₁ is a multiple of b^k − 1, it follows that

${\displaystyle N_{0}+N_{1}=b^{k}-1.}$ 

From the above,

${\displaystyle {\frac {am}{p}}}$  is an integer

Thus ${\displaystyle m\equiv 0{\pmod {p}}}$ 

And thus for ${\displaystyle k={\frac {\ell }{2}}}$ 

${\displaystyle b^{\ell /2}+1\equiv 0{\pmod {p}}}$ 

For ${\displaystyle k={\frac {\ell }{3}}}$  and is an integer

${\displaystyle b^{2\ell /3}+b^{\ell /3}+1\equiv 0{\pmod {p}}}$ 

and so on.

• Rademacher, H. and Toeplitz, O. The Enjoyment of Mathematics: Selections from Mathematics for the Amateur. Princeton, NJ: Princeton University Press, pp. 158–160, 1957. MR0081844
• 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. MR2598778

• Weisstein, Eric W. "Midy's Theorem". MathWorld.