24. N7 (GBR) A wobbly number is a positive integer whose digits in base 10 are alternately nonzero and zero, the units digit being nonzero. Determine all positive integers that do not divide any wobbly number.
24. A multiple of 10 does not divide any wobbly number. Also, if $25 \mid n$, then every multiple of $n$ ends with $25,50,75$, or 00 ; hence it is not wobbly. We now show that every other number $n$ divides some wobbly number.
(i) Let $n$ be odd and not divisible by 5 . For any $k \geq 1$ there exists $l$ such that $\left(10^{k}-1\right) n$ divides $10^{l}-1$, and thus also divides $10^{k i}-1$. Consequently, $v_{k}=\frac{10^{2}-1}{10^{2}-1}$ is divisible by $n$, and it is wobbly when $k=2$ (indeed, $v_{2}=101 \ldots 01$ ).
If $n$ is divisible by 5 , one can simply take $5 v_{2}$ instead.
(ii) Let $n$ be a power of 2 . We prove by induction on $m$ that $2^{2 m+1}$ has a wobbly multiple $w_{m}$ with exactly $m$ nonzero digits. For $m=1$, take $w_{1}=8$. Suppose that for some $m \geq 1$ there is a wobbly $w_{m}=$ $2^{2 m+1} d_{m}$. Then the numbers $a \cdot 10^{2 m}+w_{m}$ are wobbly and divisible by $2^{2 m+1}$ when $a \in\{2,4,6,8\}$. Moreover, one of these numbers is divisible by $2^{2 m+3}$. Indeed, it suffices to choose a such that $\frac{2}{2}+d_{m}$ is divisible by 4 . This proves the induction step.
(iii) Let $n=2^{m} r$, where $m \geq 1$ and $r$ is odd, $5 \nmid r$. Then $v_{2 m} w_{m}$ is wobbly and divisible by both $2^{\bar{m}}$ and $r$ (using notation from (i), $r \mid v_{2 m}$ ).