jmc

Suppose the smallest of Bill's numbers is $b$. The next few terms of the sequence are $br$, $b{r}^2$, $b{r}^3$, $b{r}^4$, and so on. Since $r$ is at least 2, $b{r}^4$ is at least $16b$. Since $16b>10b$, and $10b$ has one more digit than $b$, $16b$ has more digits than $b$, and therefore $b{r}^4$ has more digits than $b$. Since the series increases, $b{r}^5$, $b{r}^6$, and so on all have more digits than $b$. Therefore, Bill's numbers are restricted to $b$, $br$, $b{r}^2$, and $b{r}^3$; that is, he can have at most 4 numbers. An example of this is the sequence $1,2,4,8,16,\dots$, where Bill's numbers are 1, 2, 4, and 8. Hence, the largest possible value of $k$ is $\boxed{4}$.