Peter wrote: Let $S_{n}$ be the sum of the digits of $2^n$. Prove or disprove that $S_{n+1}=S_{n}$ for some positive integer $n$. Let $2^n = a_{m}10^m + a_{m-1}10^{m-1} + ... +a_{1}10 + a_0$ with $a_i$ being integers. Then $2^n \equiv a_m + a_{m-1} + ... + a_1 + a_0\equiv S(n) (mod3)$ So $S(n+1)\equiv 2^{n+1}\equiv 2.2^n\equiv 2S(n) (mod3)$ So if $S(n+1) = S(n)$ then $S(n)\equiv 0 (mod3)$ and thus $2^n\equiv 0 (mod3)$ which is impossible. Thus $S(n+1)$ cannot equal $S(n)$ for any positive integer $n$.

Analysing the above proof, we see that it shows that $S_{n+a} = S_n$ implies that a is even. While this is clearly enough to proof the impossibility of $S_{n+1} = S_n$, we might be able to deduce a little more when we work modulo 9 instead of 3. Also, since I guess that the only reason to use powers of 2 in the statement of the problem is to not give away the method of solving, the, imo, ideal proof of the impossibility of $S_{n+1} = S_n$ should (explicitly) show the following: Let $x$ be a given integer with $\gcd(x, 9) = b$. Let $S(x)$ be the sum of digits of $x$. Let $y$ be such that $S(x) = S(xy)$. Then $y \equiv 1 \bmod \displaystyle \left(\frac{9}{b}\right)$. The fact that both the proof of this and deducing the solution of our initial problem are (almost) trivial, is a good sign. Even more, since $2^a \equiv 1 \pmod 9$ implies $6 | a$, this shows that $S(2^{n+a}) = S(2^n)$ implies $6 | a$. And, since $S(2^3) = S(8) = 8 = 5 + 1 + 2 = S(512) = S(2^9)$, we *can't* do better. Btw, problem S3 (Is there a power of 2 such that it is possible to rearrange the digits giving another power of 2?) becomes an easy consequence of this. PS, in no way am I trying to criticise the proof of Winged Potato, because his might very well be the easiest and shortest proof of the theorem, I'm just trying to point out the possible merits of a proof written in a more general setting.