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Italian Mathematical Olympiad

Algebra Difficulty 4.8 AIME Find the answer Italy

Problem:

The sequence $a_{n}$ is constructed as follows: $a_{1}, a_{2}$ are integers between 1 and 9 (endpoints included); for $n \geq 3$ , if the sum of $a_{n-1}$ and $a_{n-2}$ consists of a single digit, then that sum is the value of $a_{n}$ ; if instead $a_{n-1}+a_{n-2}$ has more than one digit, the sum of its digits will be the value of $a_{n}$ (for example, if $a_{4}=7$ and $a_{5}=8$ , then $a_{6}=6$ since $7+8=15$ and $1+5=6$ ). How many possible choices of the pair $\left(a_{1}, a_{2}\right)$ are there such that $a_{2023}=9$ ?

Pick one

Solution

Solution:

The answer is $\mathbf{( C )}$ . Observe that for $n \geq 3, a_{n}$ is the number in $\{1, \ldots, 9\}$ given by the remainder of the division by 9 of $a_{n-2}+a_{n-1}$ (where 9 represents remainder 0). In particular, $a_{n-2}$ is the remainder (as above, taken in $\{1, \ldots, 9\}$ ) in the division by 9 of $a_{n}-a_{n-1}$ . Setting $a_{2023}=9$ , for each of the 9 possible choices of the value of $a_{2022}$ , the value of $a_{2021}$ is determined, and similarly all the other values of the sequence are determined: these 9 choices thus lead us to 9 possible choices of the pair $\left(a_{1}, a_{2}\right)$ .

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