2015 Countdown Problem #7: Cardinal Number Lengths

Observe that the only value of $N$ such that $f(N)=N$ is $N=4$ (since 4 is spelt as “ four ” which has 4 letters).

We first set the boundaries at $0<N≤9$ . Also, note that $\max⁡(f(N))=5$ and $f^{-1} (\max⁡{(f(N))} )=f^{-1} (5)=$ 3,7 or 8 (meaning that the longest spelt numbers from 1 to 9 are “ three ”, “ seven ” and “ eight ” at 5 letters long). With some work it can be deduced that $\lim_{k→∞}⁡{f^k (N)}=4$ for $0<N≤9$ .

We now expand the boundary to $0<N≤99$ . Observe that $\max{⁡(f(N))}=7$ when $f(N)$ is a multiple of 10 (meaning that the longest spelt multiple of 10 is “ seventy ” at 7 letters long). Hence for $0<N≤99$ , $\max⁡(f(N))=7+5=12$ and $f^{-1} (\max{⁡(f(N))} )=f^{-1} (12)=$ 77 or 78 (meaning that the longest spelt 2-digit numbers are " seventy seven " and " seventy eight " at 12 letters long). Since $f(N)≤12$ , it can be determined that $f^2 (N)≤f(12)≤6<9$ , implying that $\lim_{k→∞}⁡{f^k (N)}=4$ for $0<N≤99$ .

When we expand the boundary all the way to $0<N≤2015$ as stated in the question, it is clear that $f(N)≤99$ (meaning that any spelt number from 1 to 2015 is clearly less than 100 letters long; for observation note that $\max⁡(N)=38$ occurs when $N=1777$ ). Therefore, $\lim_{k→∞}⁡{f^k (N)}=4 \mbox{ } \forall \mbox{ } 0<N≤2015$

As such, $\sum_{N=1794}^{2015}(\lim_{k→∞}⁡{f^k (N)}) = \sum_{N=1794}^{2015}(4) =(2015-1794+1)(4)=888$