Kummer’s theorem


Given integers nm0 and a prime numberMathworldPlanetmath p, then the power of p dividing (nm) is equal to the number of carries when adding m and n-m in base p.

Proof.

For the proof we can allow of numbers in base p with leading zeros. So let

ndnd-1n0 :=n,
mdmd-1m0 :=m,

all in base p. We set r=n-m and denote the p-adic representation of r with rdrd-1r0.

We define c-1=0, and for each 0jd

cj={1for mj+rjp0otherwise. (1)

Finally, we introduce δp(n) as the sum of digits in the p-adic of n. Then it follows that the power of p dividing (nm) is

δp(m)+δp(r)-δp(n)p-1.

For each j0, we have

nj=mj+rj+cj-1-p.cj.

Then

δp(m)+δp(r)-δp(n) =k=0d(mk+rk-nk)
=k=0d((p-1)cj) +k=0d(cj-cj-1)
=k=0d(p-1)cj +cd-c-1

This gives us

δp(m)+δp(r)-δp(n)p-1=k=0dck,

the total number of carries. ∎

Title Kummer’s theorem
Canonical name KummersTheorem
Date of creation 2013-03-22 13:22:37
Last modified on 2013-03-22 13:22:37
Owner Thomas Heye (1234)
Last modified by Thomas Heye (1234)
Numerical id 14
Author Thomas Heye (1234)
Entry type Theorem
Classification msc 11A63