Is the number of nonzero digits always at least $m$? if $k,m$ is give  postive integers,for any postive integer $p$, if we define
$$p\cdot\dfrac{k^m-1}{k-1}=a_{i}k^{i}+a_{i-1}k^{i-1}+\cdots+a_{1}\cdot k+a_{0}$$
where $a_{i}\in\{0,1,2,\cdots,k-1\}$,and let set  $A=\{i|a_{i}\neq 0\}$,show that
$$|A|\ge m$$
this problem it see interesting,such  if $k=2,m=3$ I found $p=1,2,3,4,\cdots 10$ it is clear true,But How to prove this General
for example: for $k=2,m=3$,it is clear $p=2^a,a\ge 1$ is right,and we only consider $p$ is prime,
and
(1)$p=3$,then $3(2^2+2^1+2^0)=21=2^4+2^2+2^0$ so $A|=3=m$
(2)$p=5$,then $5(2^2+2^1+2^0)=35=2^5+2^1+2^0$,so $A|=3$
(3):$p=7$,then $7(2^2+2^1+2^0)=49=2^5+2^4+2^0$ so  $|A|=3$
(4):$p=11$,then $11(2^2+2^1+2^0)=77=2^6+2^3+2^2+2^0$ so  $|A|=4\ge 3=m$
(5):$p=13$,then $13(2^2+2^2+2^0)=2^6+2^4+2^3+2^1+2^0$,so $|A|=5\ge m$
(5):$p=17$,then $13(2^2+2^2+2^0)=119=2^6+2^5+2^4+2^2+2^1+2^0$,so $|A|=6\ge m$
in general: let $p=2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{k}},0\le a_{1}<a_{2}<\cdots<a_{k},a_{i}\in N$
 A: Notice that you actually claim that the number:
$$
N=N(p,k,m)=p\cdot\dfrac{k^m-1}{k-1}=p(k^0+k^1+k^2+\dots+k^{m-1})
$$
Has at least $m$ non-zero digits in number base $k$, where $a_{i}\in\{0,1,2,\cdots,k-1\}$ are its digits.

To prove the claim for $p\lt k$ is easy. Notice that for this case, we have:
$$
N=\sum_{i=0}^{m-1}p \cdot k^i
$$
Where $a_i=p\le k-1$ for $i=0,\dots,m-1$, implying $|A|=m$.
This proves the $p\lt k$ case. 

It remains to prove the $p\ge k$ case.
$(\star):$ Notice that (where $a\in\{1,2,\dots,k-1\}$ and $r\in\mathbb N$)
$$
a\cdot k^r\equiv\{a\cdot k^0,a\cdot k^1,\dots,a\cdot k^{m-1}\}\pmod{ N}
$$
Now suppose that a counter-example exists: (Proof by contradiction)
$$\begin{align}
N&=a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\\
N&\equiv a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\pmod{N}
\end{align}$$
Where $i_1,i_2,\dots,i_t$ are distinct and $t\le m-1$.
But because of $(\star)$, looking at RHS modulo $N$ we have at most (for all $p\ge k$):
$$
a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\le \sum_{i=1}^{m-1} (k-1)k^i =   k(k^{m-1}-1)\lt N
$$
Which implies $a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\not \equiv N$, which is a contradiction.
This proves the $p\ge k$ case. 

