> The other conjecture is that alternate power of two numbers

Proof:

> are divisible by 3.

Sum(i=0..N, 10^something(i)*2^i) mod 3 =

Sum(i=0..N, 2^i) mod 3 =

2^(N+1)-1 mod 3 =

0 for odd N, 1 for even N.

Q.E.D.

Peter

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[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278