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