I read What is Mathematics? by Herbert Robbins, and there is this exercise:

Consider the question of representing integers with the base $a$. In order to name the integers in this system we need words for the digits $0, 1, ..., a-1$ and for the various powers of $a$: $a, a^2, a^3, ...$. How many different number words are needed to name all numbers from zero to one thousand, for $a = 2, 3, 4, 5, ..., 15$? Which base requires the fewest?

For example, if $a = 10$, we need ten words for the digits, plus words for 10, 100, and 1000, making a total of 13. For $a = 20$, we need twenty words for the digits, plus words for 20 and 400, making a total of 22. If $a = 100$, we need 100 plus 1.

Well, I come up with the formula for the total of number words from $0$ to $x$ with base $a$: $$a + \lfloor{\log_a x}\rfloor$$

With this formula, I found that base of 4 requires the fewest number words from zero to thousand: 8.

I couldn't find any information on the web about this exercise. Can you please refer me to a more precise formula, if exists? Can you please surface any issue with it? Is that true at all?

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    $\begingroup$ I think you've solved the problem in the natural way. I doubt there's a better formula. +1 for reading Robbins and thinking about problems. $\endgroup$ – Ethan Bolker Aug 31 '17 at 13:13
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    $\begingroup$ The mathematical system of simply writing the digits in place notation and allowing that to imply the relevant power suggests that the need for words like "hundred" could be sidestepped. How many days in a year ? "three-six-five". For clarity perhaps an end-of-number word could be added. $\endgroup$ – Joffan Aug 31 '17 at 13:46
  • $\begingroup$ @Joffan I guess it won't be efficient in longer numbers, like 6,000,005. $\endgroup$ – SRachamim Sep 1 '17 at 23:25

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