If k is any divisor of the period of the decimal expansion of a/p, 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, 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
Midy's theorem and its extension do not depend on special properties of the decimal expansion, but work equally well in any baseb, provided we replace 10k − 1 with bk − 1 and carry out addition in base b. For example, in octal In duodecimal
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 where N is the integer whose expansion in base b is the stringa1a2...aℓ. Note that bℓ − 1 is a multiple of p because 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 bℓ − 1 is a multiple of bk − 1. Say bℓ − 1 = m, so But bℓ − 1 is a multiple of p; bk − 1 is not a multiple of p ; and p is a prime; so m must be a multiple of p and is an integer. In other words, 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 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 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 N0 and N1cannot both equal 0 and cannot both equal bk − 1, so and since N0 + N1 is a multiple of bk − 1, it follows that
Corollary
From the above, Thus And thus for For and is an integer and so on.