32. Find the number of ordered 7-tuples of positive integers $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}\right)$ that have both of the following properties:
(i) $a_{n}+a_{n+1}=a_{n+2}$ for $1 \leq n \leq 5$, and
(ii) $a_{6}=2005$.
32. Ans: 133
From (i), it is easy to see that the whole ordered 7-tuple is determined by $a_{1}$ and $a_{2}$. It is easy to check that $a_{6}=3 a_{1}+5 a_{2}$. Thus by (ii), we have $3 a_{1}+5 a_{2}=2005$. It follows that $a_{1}$ is divisible by 5 . Write $a_{1}=5 k$. Then we have $15 k+5 a_{2}=$ $2005 \Longrightarrow 3 k+a_{2}=401$. The possible values of $a_{2}$ are $\{2,5,8, \cdots, 398\}$, if both $k$
and $a_{2}$ are positive. Note that each of the above values of $a_{2}$ determines a unique positive integral value of $a_{1}$ satisfying the equation $3 a_{1}+5 a_{2}=2005$. Therefore, there are 133 such sequences.