Take $x \in \Bbb N$, $x \le9$ and $m \in \Bbb N$. Now we define a function $d_s(n): \Bbb N \to \Bbb N$ as the digit sum of $n$ in base $10$. Now let's say we have a lower bound $b_l$ and an upper bound $b_u$ (both implicitly as functions of $m$) so that $b_l \le d_s(x^m) \le b_u$. Find the values $b_l$ and $b_u$.

The only things i have come up with are $b_u \le 9 \lfloor m*log_{10}(x)+1 \rfloor$ since that is digit sum of an integer with the same number of digits, all of them having a value of $9$ and $b_l \ge 1$ for obvious reasons. I can't seem to find a way to approach either of these values other than the trivial remarks I made earlier. I used Mathematica to evaluate some values and there does seem to be a pattern.

  • $\begingroup$ You can show that $b_l\gt 1$ for similarly obvious reasons (what are the numbers $n$ with $d(n)=1$?) and using similar techniques it should be pretty straightforward to push past this to $b_l\gt 2$ and even to $b_l\gt 3$. Beyond that, though, things get substantially hairy. $\endgroup$ – Steven Stadnicki Jul 15 '15 at 21:08
  • $\begingroup$ (Also, I've tweaked your notation a bit, adding a parameter where one was needed and changing your exponent variable from $l$ to $m$ to avoid confusion with $b_l$; please feel free to refine as you see fit) $\endgroup$ – Steven Stadnicki Jul 15 '15 at 21:13
  • $\begingroup$ If x=1 then $b_l$ can equal 1, but ofcourse x=1 is trivial. $\endgroup$ – Frank Jul 15 '15 at 21:13
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    $\begingroup$ @Frank: You may want to see this MO question. $\endgroup$ – mathlove Jul 15 '15 at 21:33

Here's the rough outline of the argument I mentioned in the comments that $b_l\gt 3$ in all non-trivial cases: Suppose $d(x^m)=1$ for some $m$. Then this implies that $x^m=10^i$ (since the only numbers with $d(n)=1$ are powers of $10$); but then it must be the case that $5|x$ and therefore $x=5$ (since $x\leq 9$), but clearly powers of $5$ aren't powers of $10$ (for one thing, they're odd).

Now, if $d(x^m)=2$ then things break down into two cases: $x^m=2\cdot 10^i$ for some $i$, or $x^m=10^i+10^j$. The first case can be handled using the same argument that $5|x$; this also forces $j=0$ in the second case, so that we're left with $x^m=10^i+1$. Now, if this is ever to be the case then we must have $x$ odd (since $2\not\mid 10^i+1$); $x=3$ and $x=9$ are impossible since $10^i+1\not\equiv0\pmod 3$, and similarly $x=5$ is impossible since $10^i+1\not\equiv 0\pmod 5$, so the only possibility is $x=7$. But now if $10^i\equiv-1\pmod 7$ then $i\equiv 3\pmod 6$ (you should be able to show this easily) and in particular $i$ is odd — but this implies that $11|10^i+1$ (since then $10^i\equiv (-1)^i\equiv -1\pmod {11}$) and so it's impossible that $10^i+1\equiv 7^m$.

Finally, if $d(x^m)=3$ then $3|x^m$ and so $3|x$ — in which case, $3^m|x^m$ and in particular $9|x^m$ (except for the trivial case $m=1$). But of course, if $9|x^m$ then $9|d(x^m)$, giving the needed contradiction.


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