Midy's theorem

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' 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 052631+578947+368421=999999. Similarly, dividing the repeating portion into 3-digit numbers and summing them gives 052+631+578+947+368+421=2997=3\times999. ==Midy's theorem in other bases==

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 \begin{align} & \frac{1}{19}=0.\overline{032745}_8 \\[8pt] & 032_8+745_8=777_8 \\[8pt] & 03_8+27_8+45_8=77_8. \end{align} In dozenal (using inverted two and three for ten and eleven, respectively) : \begin{align} & \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{align} ==Proof of Midy's theorem==

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 \begin{align} & \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} \end{align} where N is the integer whose expansion in base b is the string a1a2...aℓ. 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 bk − 1. (To see this, substitute x for bk; then bℓ = xh and x − 1 is a factor of xh − 1. ) Say b ℓ − 1 = m(bk − 1), so \frac{a}{p}=\frac{N}{m(b^k-1)}. But b ℓ − 1 is a multiple of p; bk − 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 \frac{am}{p}=\frac{N}{b^k-1} is an integer. In other words, N\equiv0\pmod{b^k-1}. Now split the string a1a2...aℓ into h equal parts of length k, and let these represent the integers N0...Nh − 1 in base b, so that \begin{align} 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{align} 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 N=\sum_{i=0}^{h-1}N_ib^{ik}=\sum_{i=0}^{h-1}N_i(b^{k})^i \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 and since N0 + N1 is a multiple of bk − 1, it follows that N_0+N_1 = b^k-1. ==Corollary==

From the above, \frac{am}{p} is an integer Thus m \equiv 0 \pmod p And thus for k = \frac{\ell}{2} b^{\ell/2}+ 1 \equiv 0 \pmod p For k = \frac{\ell}{3} and is an integer b^{2\ell/3} + b^{\ell/3} + 1 \equiv 0 \pmod p and so on. ==Notes==