5. Let $a_{0}, a_{1}, a_{2}, \ldots$ be integers such that $a_{0}=19, a_{1}=25$, and for every $n \geq 0$ we have $a_{n+2}=2 a_{n+1}-a_{n}$. What is the smallest $i>0$ for which $a_{i}$ is a multiple of 19?
(A) 19
(B) 25
(C) 38
(D) 44
(E) 50.
5. The answer is (A). The recurrence rule given in the text is equivalent to saying $a_{k+1}-a_{k}=a_{k}-a_{k-1}=\cdots=a_{1}-a_{0}=d$ (the difference between successive terms remains constant), and thus one recognizes in $a_{k}$ an arithmetic progression. It can therefore be derived that $a_{k}=19+6k$ for every $k$ (a formal proof of this result can be conducted by induction). For this number to be a multiple of 19, it is necessary and sufficient that $6k$ be a multiple of 19, and this occurs for the first time when $k=19$.