10.17 A sequence is recursively defined as: $t_{1}=1$, for $n>1$, when $n$ is even, $t_{n}=1+t \frac{n}{2}$; when $n$ is odd, $t_{n}=\frac{1}{t_{n-1}}$. If it is known that $t_{n}=\frac{19}{87}$, then the sum of the digits of $n$ is 
(A) 15.
(B) 17.
(C) 19.
(D) 21.
(E) 23.
(38th American High School Mathematics Examination, 1987)
［Solution］We use the reverse method (backtracking).
From the problem, we know that when $n$ is even, $t_{n}=1+t \frac{n}{2}>1$, thus for odd terms, $t_{n}=\frac{1}{t_{n-1}}<1$. Therefore, $\frac{19}{87}$ is an odd term. Its previous term is $\frac{87}{19}$. Note that
$$
\frac{87}{19} \rightarrow 1+\frac{68}{19} \rightarrow 1+1+\frac{49}{19} \rightarrow 1+1+1+\frac{30}{19} \rightarrow 1+1+1+1+\frac{11}{19} ;
$$
$\frac{11}{19}$ is an odd term, so its previous term is $\frac{19}{11}$. Note that
$$
\begin{array}{l}
\frac{19}{11} \rightarrow 1+\frac{8}{11}, \\
\frac{11}{8} \rightarrow 1+\frac{3}{8}, \\
\frac{8}{3} \rightarrow 1+\frac{5}{3} \rightarrow 1+1+\frac{2}{3}, \\
\frac{3}{2} \rightarrow 1+\frac{1}{2} .
\end{array}
$$

And $\frac{1}{2}$ is exactly the 3rd term. We backtrack to know:
\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
\hline Number $s_{n}$ & $\frac{1}{2}$ & $\frac{3}{2}$ & $\frac{2}{3}$ & $\frac{5}{3}$ & $\frac{8}{3}$ & $\frac{3}{8}$ & $\frac{11}{8}$ & $\frac{8}{11}$ & $\frac{19}{11}$ & $\frac{11}{19}$ & $\frac{30}{19}$ & $\frac{49}{19}$ & $\frac{68}{19}$ & $\frac{87}{19}$ & $\frac{19}{87}$ \\
\hline Term number $n$ & 3 & 6 & 7 & 14 & 28 & 29 & 58 & 59 & 118 & 119 & 238 & 476 & 952 & 1904 & 1905 \\
\hline
\end{tabular}
$\therefore \quad n=1905$, the sum of its digits is 15.
Therefore, the answer is $(A)$.