# 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. This can be proven by induction. 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

Instead of setting $a_1$ equal to something and $a_2$ equal to something, note that it is rather easier to set $a_1=x^2$ and $a_3=y^2$ so that $a_2=xy,a_4=y(2y-x),a_5=(2y-x)^2$ and so on until you reached $a_{13}=(6y-5x)^2$ (Or simply notice the pattern), so $6y-5x=\sqrt{2016}=12\sqrt{14}$. Note that since each of the terms has degree 2 so if you multiply $x$ and $y$ by $\sqrt{14}$ you multiply each term by $14$ so each term is still a integer if the terms are already integers before you multiply $x$ and $y$ by $\sqrt{14}$, so let $w=\frac{x}{\sqrt{14}}$ and $z=\frac{y}{\sqrt{14}}$ so $6z-5w=12$. Then, for the sequence to be strictly increasing positive integers we have $(w,z)=(6,7)$ so $x=6\sqrt{14}$ and $a_1=x^2=6^2 \cdot 14=\boxed{504}$~Ddk001

## Solution 4(very risky and very stupid)

The thirteenth term of the sequence is $2016$, which makes that fourteenth term of the sequence $2016+r$ and the $15^{\text{th}}$ term $\frac{(2016+r)^2}{2016}$. We note that $r$ is an integer so that means $\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$, which is $168$. We bash all the way back to the first term and get our answer of $\boxed{504}$.

## Solution 5

Let $a_{2k-1}=s$ where $k=1$. Then, $a_{2k}=sr \Longrightarrow a_{2k+1}=sr^2$. Continuing on, we get $a_{2k+2}=sr(r-1)+sr^2=sr(2r-1) \Longrightarrow a_{2k+3}=sr^2(\frac{2r-1}{r})^2=s(2r-1)^2$. Moreover, $a_{2k+4}=s(2r-1)(r-1)+s(2r-1)^2=s(2r-1)(3r-2) \Longrightarrow a_{2k+5}=s(2r-1)^2(\frac{3r-2}{2r-1})^2=s(3r-2)^2$.

It is clear now that $a_{2k+2c}=s(cr-(c-1))((c+1)r-c)$ and $a_{2k+2c-1}=s(cr-(c-1))^2$. Plugging in $c=6$, $a_{13}=s(6r-5)^2=2016$. The prime factorization of $2016=2^5\cdot3^2\cdot7$ so we look for perfect squares.

$6r-5\equiv (6r-5)^2\equiv 1\pmod{6}$ if $r$ is an integer, and $\frac{\omega+5}{6}=r \Longrightarrow 6\mid{s}$ if $r$ is not an integer and $\omega$ is rational, so $6\mid{s}$. This forces $s=2\cdot3^2\cdot7\cdot{N}$. Assuming $(6r-5)$ is an integer, it can only be $2^x$, $x\in{1,2}$.

If $6r-5=2^1$, $r=\frac{7}{6}$. If $6r-5=2^2$, $r=\frac{3}{2}$. Note that the latter cannot work since $a_{2k+1}=s(\frac{9}{4}) \Longrightarrow 4\mid{s}$ but $N=1 \Longrightarrow s=2\cdot3^2\cdot7$ in this scenario. Therefore, $r=\frac{7}{6} \Longrightarrow s=\frac{2016}{2^2}=504$. Plugging back $k=1$, $a_1=s=\boxed{504}$

Bash all the way from $a_1$ and $a_2$ to $a_{13}=2016$ to get to the equation $a_1\left(\frac{6a_2}{a_1}-5\right)^2=2016$ and then $a_2=2\sqrt{14a_1}+\frac{5}{6}a_1$ (direct consequence). Notice that from the two equations we can deduce that:

1) $a_1\mid2016$,

2) $6\mid a_1$, and

3) $\sqrt{\frac{a_1}{14}}\in\mathbb{Z}$.

Since $2016=2^5\cdot3^2\cdot7=14(2^4\cdot 3^2)$, the only possible values of $a_1$ that satisfy the conditions are $126$, $504$, and $2016$ itself. Obviously $2016$ cannot be a correct answer choice (from AIME rules), and if $a_1=126$, $a_3\not\in\mathbb{Z}$, so $\boxed{504}$ is the solution.

~eevee9406

## Video Solution

~MathProblemSolvingSkills.com