Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $n\in\mathbb{N}^*$ we note $s(n)$ the sum of the digits of $2^n$. For example $s(4)=7$ because $2^4=16$. Is it possible to find an integer $n$ such that $s(n)=s(n+1)$?

share|cite|improve this question
all $n \ge 0$ when in binary ;) – ratchet freak Oct 12 '11 at 15:06
up vote 26 down vote accepted

No, because the digit sum for $k$ is congruent to $k$ modulo 3, and $k$ and $2k$ are only congruent mod $3$ if $k$ is divisible by $3$.

share|cite|improve this answer
Using the standard casting out $9$s, we have same thing mod $9$. (+1) – robjohn Oct 12 '11 at 4:37

HINT $\ $ More generally, just as in casting nines, by casting $\rm\:b-1$'s in radix $\rm\:b\:$ we conclude that if $\rm\ 2^{n+1}\:$ and $\rm\:2^{n}\:$ have equal radix $\rm\:b\:$ digit sums then $\rm\: b-1\ |\ 2^{n+1}-2^n = 2^n\:,\:$ hence $\rm\:b = 2^k+1\:.$

For example, in radix $5$ note that $\ 2^2 = 4,\ 2^3 = 13,\ 2^4 = 31\:$ all have digit sum $\:= 4\:,\:$ and, similarly, in radix $9\!:\ \:2^3 = 8,\ 2^4 = 17,\ 2^5 = 35,\ 2^6 = 71,\ 2^7 = 152\ $ all have digit sum $\:= 8\:.$

Even more generally, a result analogous to that above remains true if we replace $2$ by any prime $\rm\:p\:$ such that $\rm\: \gcd(p-1,b-1) = 1\:.$

share|cite|improve this answer
Thanks for the example. At first, I thought you were saying something else, until I thought more about it. (+1) – robjohn Oct 12 '11 at 4:48

No: $2 \equiv -1 \pmod 3$, so $2^n \equiv (-1)^n \pmod 3$, and therefore the sequence $\langle 2^n:n\in\mathbb{N}^*\rangle$ is $\langle -1,1,-1,1,\dots\rangle$ when reduced mod $3$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.