This week’s GeekDad Puzzle of the Week is based around the number “1” (one) and the number of times the number appears in the set of counting numbers. In the set of numbers from 1 to ten, the digit “1” appears twice: once on the “1”, and then again on the “10.” If we keep a running total, we can see that there are 12 “1s” present counting from 1 to 20.
| 1 | 1 | 11 | 4 | |
| 2 | 1 | 12 | 5 | |
| 3 | 1 | 13 | 6 | |
| 4 | 1 | 14 | 7 | |
| 5 | 1 | 15 | 8 | |
| 6 | 1 | 16 | 9 | |
| 7 | 1 | 17 | 10 | |
| 8 | 1 | 18 | 11 | |
| 9 | 1 | 19 | 12 | |
| 10 | 2 | 20 | 12 |
For the most part, it seems like the running count of observed 1s is far lower than the number up to which we have counted at that point; at 100, we have only seen 21 ones, and at 1000 only 301.
Is this true for all counting numbers under 10 million? That is, are there runs of observable counts that equal or exceed the number up to which we have counted at that point? If so, what are they? If not, why is it not possible?
As always, please submit your responses to GeekDad Central. If your response if correct (or reasonably well reasoned) it will be entered into a random drawing of all such responses. This week’s prize is a $50 Gift Certificate from the fine folks at ThinkGeek.
Happy puzzling!
