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

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 $504.$

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 $\displaystyle \frac{(2016+r)^2}{2016}$ . We note that $r$ is an integer so that means $\displaystyle \frac{r^2}{2016}$ 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

2016 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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