Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 3.8 AMC 10/12 Find the answer Italy

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 $19$ ?

Pick one

Solution

Solution:

The answer is (A). The recurrence rule given in the statement is equivalent to saying $a_{k+1}-a_{k}=a_{k}-a_{k-1}=\cdots=a_{1}-a_{0}=d$ (the difference between two consecutive terms remains constant), and so we recognize $a_{k}$ as an arithmetic progression. We can then derive that $a_{k}=19+6 k$ for every $k$ (a formal proof of this result can be carried out by induction). For this number to be a multiple of $19$ , it is necessary and sufficient that $6 k$ be a multiple of $19$ , and this happens for the first time when $k=19$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.