Midy's theorem

Extended Midy's theorem

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 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,

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 base b, provided we replace 10^k − 1 with b^k − 1 and carry out addition in base b.

For example, in octal

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

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

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

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

is an integer. In other words,

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

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

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

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

Corollary

From the above,

Thus m = 0 ( mod p )

And thus for k = l 2

For k = l 3 and is an integer

and so on.