Is the estimation of number's name's length and comma-grouping feasible?

I am thinking in a mathematical problem that probably is already formulated and even solved. It is about big integers and someting else.

Let n be an integer positive number.

• For n := 1,000 we have written words to name it: “thousand” in English, “mil” in Spanish, “mille” in French, “Tausend” in German, and “sen” or “ban” in Japanese (written in latin alphabet).
• For n := 1,000,000 we have “million”.
• For n := 1E9 (“E” is read “times ten raised to the power of” and the number, as everybody knows) many people will call it “billion”, but in most european countries people will say “thousand million”, and call 1E12 “billion”, while is somehow more common to hear “trillion”.

Suppose now that we have a very big number, for instance: n := 3.1416E9999, which is 31416 followed by 9995 zeroes, if I'm not wrong. Questions:

1. If we were to actually write this rather unfathomable number, and separate in “groups” of three digits as is the custom (counting from the “right” hand side of the number, or the “end” if you prefer), and marking this groups with a comma, then:

1. How many commas would there be?
2. Where would go the number's first comma?
• 3,141,600...
• 31,416,000...
• 314,160,000...
2. What function could give us the length of the word used to pronounce the whole number? And what is the length of word used to say the leftmost part, the largest “unit” or “group”?

I think that the first question is a straight induction problem, but it's too tricky for me. And about the second one, I am no sure if I can solve it. First of all, it may be impossible to compose a single word with so many Greek or Latin prefixes.

I've been looking up the CPAN for a perl module that could answer my question, but my research was unsuccessful. I want to know what is the length of that prefix (maybe even in syllables rather than in letters) in order to calculate how much time would it take to pronounce the number. This should be an estimate because humans usually stop to breathe when talking, and there's also the need to drink water and sleep, and some words take much time to pronunce them than others. And why do I want to know this? Because I want to know exactly when a number is pronounceable and unpronounceable.

I don't know if these results can be useful or simply trivial, but if the mathematical science cannot answer this, I think we are doomed.

Please, spare me for my grammatical mistakes, since English is not my first language.

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Uhm, I'm from The Netherlands and we don't say "a thousand million" we say "miljard", the Dutch word for "billion". Germans say "billion" or "milliarde" and so do the French. –  Stijn Mar 24 '11 at 18:17
Mathematical science can solve this, don't worry. But from a mathematical point of view there's nothing exciting here. This shouldn't deter other would-be answerers. –  Yuval Filmus Mar 24 '11 at 18:25
@Stijn I'm Spanish and we say billón and billones for 1E12 and mil millones or miles de millones for 1E9. It's possible to use millardos but it's extremely rare, nobody say or write that normally. Maybe German related languages are prone to use the “sort-scale” and Latin related languages are to the “long-scale”. en.wikipedia.org/wiki/Long_and_short_scales –  user8664 Mar 24 '11 at 18:43
There is a very detailed discussion of the naming of large numbers in Conway and Guy's "The Book of Numbers". books.google.ca/… –  Byron Schmuland Mar 24 '11 at 19:03
The word for thousand in Japanese is 千 (sen), and the word for ten thousand in 万 (man), but there is no word ban as far as I know. –  Zhen Lin Apr 27 '11 at 9:45

1 Answer

Say that our number is $a \mathrm E n$, or in other words $a\times 10^n$ where $1<a<10$. Also suppose $a$ is a rational number with less than $n$ decimal digits. (So that $a\times 10^n$ is an integer) For example, in the above case, $a=3.1416$ and $n=9999$, and $a$ certainly has less than $n$ digits.

1: The number of commas will be: $$\left\lfloor \frac{n-1}{3}\right\rfloor.$$

2: To find where the first comma goes, look at the remainder of $n$ after division by 3, call this $r$. If the remainder is one, the first comma goes one to the right of the first number. If the remainder is two, the first comma goes two to the right of the first number. Lastly if the remainder is 0 (that is, $n$ is divisible by 3) the first comma goes three to the right of the first number.

So in our case, since $n=9999$ is divisible by 3, we see that the first comma goes three to the right, and hence we would write: 314, 160, 000, ...

3: It is not clear how long the word could be, but we can put a makeshift minimum number of letters. Since there are 26 characters in the alphabet, and we need a word to represent every 3 zeros, the "word" representing $\left(10^3\right)^n$ would require at least $$\left\lceil \frac{\log n}{\log 26}\right\rceil$$ letters.

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@eric_naslund You wrote that “If the remainder is two, the first comma goes one to the right of the first number”, but, is it possible that you meant the two to the right? –  user8664 Mar 24 '11 at 18:50
@Michael: Big mistake there! Fixed now, thanks! –  Eric Naslund Mar 24 '11 at 18:51