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I was playing with the letters in numbers written in English and I found something quite funny. I found that if you count the number of letters in the number and write this as a number and then count the number of letters in this new number and keep repeating the process, you will arrive at the number 4. I've confirmed this (using a computer program) for all numbers up to 999999 and was wondering if there's a way to prove this or to find a counter example for which it does not hold.

Just to give an example of the above statement, let's start with thirty seven (I chose this randomly) Thirty seven has 11 letters in it, Eleven has 6 letters in it, Six has three letters in it, Three has 5 letters in it, Five has 4 letters in it. It may look like I just picked this number, so let me show this for another random number, say 999. Nine hundred and ninety nine has 24 letters in it, Twenty four has 10 letters in it, Ten has 3 letters in it, Three has 5 letters in it, Five has 4 letters in it.

What are your thoughts on how to prove this?

(Just a note: I only confirmed this for numbers written in the standard British way of writing numbers - for example 101 is one hundred and one)

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3 Answers 3

up vote 17 down vote accepted

Define $f: \mathbb{N} \to \mathbb{N}$ as the number of letters in a given natural number spelled out.

Four is the only fixed point under $f$, and it's not too difficult to see that $f$ is almost always strictly decreasing with the only exceptions being one, two, three and four. So the $n^{th}$ iterate of $f$ must eventually become smaller than 5, which doesn't leave very many cases to verify.

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Is there actually a systematic way to name arbitrarily large numbers in English? It seems that one would have to specify this before claiming that your $f$ is eventually decreasing. –  user83827 Dec 11 '11 at 19:56
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@ccc I suppose one would have to make the assumption that the numbering system is relatively uniform - and if no such nomenclature exists for a certain number, that the number of letters in that number would not spike by an enormous amount of digits. Perhaps one could adopt the convention of saying one million million for one million squared, for instance, so that all numbers may be accounted for. –  analysisj Dec 11 '11 at 20:01
    
@ccc Do those Hebrew letters used for Cantor algebra accomplish that (naming arbitrarily large numbers)? I don't recall exactly though... –  Ellie Kesselman Dec 11 '11 at 20:06
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@ccc This is a bad question: Why have you NEVER voted on any question, up or down? (That needn't be answered.) –  Ellie Kesselman Dec 11 '11 at 20:10
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There is such a systematic method of naming numbers in English, although the details escape me. Given that a 1000-factor increase in the size of the number results in roughly a 28-35 letter increase in name size ("three hundred and thirty three -illion" and so on) it's clear that the naming convention will never catch up. Consider also what it would take for the size of the name to grow as the number: you'd have to write in unary. –  jprete Dec 11 '11 at 22:11

Note that the number of letters in the number is almost always going to be less than the number itself; in fact, this should be true for all numbers greater than four. Four is the only number which has this property (that the number of letters is equal to the number itself). Therefore, we can say that if a number repeats eventually, it must repeat at 4. Furthermore, since the value of the number of letters in the number is always less than the number itself for values greater than 4, the number is always decreasing until four.

So, all that remains is to show that the numbers 3,2, and 1 always go to 4 eventually, and then we can know that it will repeat. 3 goes to 5 which goes to 4. Both 1 and 2 go to 3 which go to 5 which go to 4. Though this is fairly informal, this is the gist of how it could be shown.

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As an interesting side note, no such number should occur in French. There exists no number when spelled out such that the number of letters is equal to the number, but you can get patterns such as 3,5,4,6,3,5,4,6,... because 3 is trois which goes to 5 which is cinq which goes to 4 which is quatre which goes to 6 which is six, which goes back to three... –  analysisj Dec 11 '11 at 19:50

HINT

$2 \to 3 \to 5 \to 4$

$3 \to 5 \to 4$

$4 \to 4$

$5 \to 4$

$6 \to 3 \to 5 \to 4$

$7 \to 5 \to 4$

$8 \to 5 \to 4$

$9 \to 4$

$10 \to 3 \to 5 \to 4$

etc.

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I don't mean this as an exhaustive list (!), as this is what you have shown for n < 999999. Rather, if you can show that the function (as Dustan writes) decreases to eventually give you a value less than 10 (or 5), then you're done. –  The Chaz 2.0 Dec 11 '11 at 19:41
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You should edit your comment into your answer –  Henry Dec 12 '11 at 0:05

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