Midy's theorem

In mathematics, Midy's theorem, named after French mathematician E. Midy, 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 the OEIS). If the period of the decimal representation of a/p is 2n, so that a p = 0. a 1 a 2 a 3 … a n a n + 1 … a 2 n ¯ {\displaystyle {\frac {a}{p}}=0.{\overline {a_{1}a_{2}a_{3}\dots a_{n}a_{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 {\displaystyle a_{i}+a_{i+n}=9} a 1 … a n + a n + 1 … a 2 n = 10 n − 1. {\displaystyle a_{1}\dots a_{n}+a_{n+1}\dots a_{2n}=10^{n}-1.} For example, 1 13 = 0. 076923 ¯ and 076 + 923 = 999. {\displaystyle {\frac {1}{13}}=0.{\overline {076923}}{\text{ and }}076+923=999.} 1 17 = 0. 0588235294117647 ¯ and 05882352 + 94117647 = 99999999. {\displaystyle {\frac {1}{17}}=0.{\overline {0588235294117647}}{\text{ and }}05882352+94117647=99999999.} 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 10 − 1. For example, 1 19 = 0. 052631578947368421 ¯ {\displaystyle {\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. {\displaystyle 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 × 999. {\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 − 1 with b − 1 and carry out addition in base b. For example, in octal 1 19 = 0.