# Difference between revisions of "2016 AIME I Problems/Problem 10"

## Problem

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 1

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 in the sequence don't have enough elements before them to go all the way back to $a_1$ while still staying as positive integers. $a_{13}=2016=14\cdot 144$, so $a_1=14\cdot 36=\fbox{504}$.

~IYN~

## Solution 2

Setting $a_1 = a$ and $a_2 = ka$, the sequence becomes:

$$a, ka, k^2a, k(2k-1)a, (2k-1)^2a, (2k-1)(3k-2)a, (3k-2)^2a, \cdots$$ and so forth, with $a_{2n+1} = (nk-(n-1))^2a$. Then, $a_{13} = (6k-5)^2a = 2016$. Keep in mind, $k$ need not be an integer, only $k^2a, (k+1)^2a,$ etc. does. $2016 = 2^5*3^2*7$, so only the squares $1, 4, 9, 16, 36,$ and $144$ are plausible for $(6k-5)^2$. But when that is anything other than $2$, $k^2a$ is not an integer. Therefore, $a = 2016/2^2 = 504$.

## Solution 3

This is not a hard bash. You can try the ratios $\frac 2 3$, $\frac 3 4$, and $\frac {11} {12}.$ Working backwards from $\frac {11} {12},$ we get $504.$