Interesting Property of Numbers in English 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)
 A: 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.
A: 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. 
A: 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.
