Multiples and decimal expansion... If $d \in \{1;\cdots;9\}$ and $x \in \mathbb{N^*}$, $N_d(x)$ denotes the number of digits equal to $d$ in the decimal expansion of $x$.

If $n$ and $m$ are in $\mathbb{N^*}$, is there $a$ in $\mathbb{N}^*$ such that : $\forall d \in \{1;\cdots;9\}$ , $N_d(an)=N_d(am)$ ?
Not completely rigorous, but let $m$ be the concatenation of to copies of $n$.  Then (ignoring carries at the joint) there will be twice as many of each digit in $am$ as there are in $an$
Generally no. For example $N_0(a \times 1) &lt; N_0(a \times 1000)$ for  $a \in \mathbb N^* $.
But this is not counterexample.