Solution to Problem (1) by Gareth McCaughan:

I accepted this solution after some thought:

`The conjecture on your web page about taking differences of digits is true.Suppose to begin with that all digits are 0 or 1. Then aftert iterations you have, mod 2, a_new(k) = sum {0..t} of (t choose r)times a(k+r mod n).1. If n is a power of 2, then take t=n and note that all thosebinomial coefficients are even other than (n choose 0)and (n choose n), and that those two catch the same a(k),so to speak; so all the a_new(k) are 0. That was allassuming that the digits are 0 or 1, but notice that the sameargument works with the (digits mod 2), so we've provedthat after iterating n times all your digits are even. So do itagain and get all the digits to be multiples of 4, then againmaking them multiples of 8, and so on until you reach apower of 2 larger than any of the original digits, at whichpoint all the digits must equal 0. So if n is a power of 2,you must end up with the all-zero string.2. If n is odd and > 1, start with any string of 0s and 1s other thanall-0 or all-1. I claim you never end up at all-0. Why? Becausethe only possible predecessors of all-0 are all-0 and all-1, andthere is *no* possible predecessor of all-1 (easy exercise forthe reader). So when n is odd and > 1, there must be some possiblefate other than all-0.3. If n is neither odd nor a power of 2, write it as m*2^k wherem is odd. Take any string S of 0s and 1s of length m, other thanall-0 and all-1, and repeat it 2^k times. Now the t'th iterate ofthis string is the t'th iterate of S, repeated 2^k times (easyexercise for the reader), and *that* is never all-0. So whenn is an odd number > 1 times a power of 2, there must be somepossible fate other than all-0. We're done.                        *Back to Maths Corner`