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, $a = 2016/2^2 = 504$ .

Thanks for reading, Rowechen Zhong.

Solution 3(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 $\boxed{504}$ .

-Pleaseletmewin

Solution 4

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 $k=1$ , $a_1=s=\boxed{504}$

Video Solution

https://youtu.be/fVRHaPE88Cg

~MathProblemSolvingSkills.com

2016 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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