Set r:=n, q:=k ((...,k1,k0)p(...,n1,n0)p)≡((...,k2,k1)p(...,n2,n1)p)(k0n0)≡...≡∏(kini) Pascaltriangle mod So when is (kn) odd? (kn)=∏(kn)mod2 Criterion for you parity of (kn) (kn) is odd ⇐> binaryform of n dominates binaryform of k.