2016 AIME I Problems/Problem 10

A strictly increasing sequence of positive integers $a_1$ , $a_2$ , $a_3$ , $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ , $a_{2k}$ , $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ , $a_{2k+1}$ , $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ .

Solution

We first create a similar sequence where $a_1=1$ and $a_2=2$ . Continuing the sequence,

$1, 2,4,6,9,12,16,20,25,30,36,42,49,\cdots$

Here we can see a pattern; every second term (starting from the first) is a square, and every second term (starting from the third) is the end of a geometric sequence. Similarly, $a_{13}$ would also need to be the end of a geometric sequence (divisible by a square). We see that $2016$ is $2^5 \cdot 3^2 \cdot 7$ , so the squares that would fit in $2016$ are $1^2=1$ , $2^2=4$ , $3^2=9$ , $2^4=16$ , $2^2 \cdot 3^2 = 36$ , and $2^4 \cdot 3^2 = 144$ . By simple inspection $144$ is the only plausible square, since the other squares don't have enough numbers before them to go all the way back to $a_1$ . $a_{13}=2016=14\cdot 144$ , so $a_1=14\cdot 36=\fbox{504}$ .

~Iy31n~

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2016 AIME I Problems/Problem 10

Solution