Mock AIME II 2012 Problems/Problem 2

Problem

Let $\{a_n\}$ be a recursion defined such that $a_1=1, a_2=20$ , and $a_n=\sqrt{\left| a_{n-1}^2-a_{n-2}^2 \right|}$ where $n\ge 3$ , and $n$ is an integer. If $a_m=k$ for $k$ being a positive integer greater than $1$ and $m$ being a positive integer greater than 2, find the smallest possible value of $m+k$ .

Solution

The sequence goes as follows: $1, 20, \sqrt{399}, 1, \sqrt{398}, \sqrt{397}, 1, \sqrt{396}, \sqrt{395}, 1, \cdots$ Note that this sequence repeats, because once we get to $1$ , we subtract $1$ from the number under the square root sign and take the square root of that, then when you take the difference of the squares of these two numbers, they have a difference of $1$ , so you get back to $1$ .

The first perfect square less than $20^2$ is $19^2=361$ . Therefore $k=\sqrt{361}=19$ .

We present two ways to finding $m$ :

Way 1: Now notice that, for $n\equiv1(\bmod3)$ , we have $a_n=1$ . Also, the terms under the radical counts down one for every $a_n$ such that $a_n\not\equiv1(\bmod3 )$ . The number of $n\equiv1(\bmod 3)$ before a number $n$ can be seen to be $\lceil \frac{n}{3}\rceil$ . Therefore, we can see that $a_n=\sqrt{401-n+\lceil\frac{n}{3}\rceil}$ . Setting this equal to $19$ , we see that $n-\lceil\frac{n}{3}\rceil=40$ . We can see that both $n=60$ and $n=61$ satisfy this, however, since $61\equiv1(\bmod3)$ , we disregard this, and we find that $a_{60}=19$ , so $m=60$ .

Way 2: Look at the sequence by grouping terms as follows: $1, (20, \sqrt{399}, 1), (\sqrt{398}, \sqrt{397}, 1), (\sqrt{396}, \sqrt{395}, 1), \cdots (\sqrt{362}, \sqrt{361}, 1)$ . Note that the number of these groups of three that we have is equal to the number of terms in the sequence $(361, 363, 365, \cdots 399)$ . This has $\frac{399-359}{2}=20$ terms in it. Before the last group, we have $19*3+1=58$ terms, and $\sqrt{361}$ is therefore the $60$ th term, and $m=60$ .

Both ways lead to $m=60$ and $k=19$ , so the answer is $60+19=\boxed{079}$ .

To show that no smaller sums can be created, note that the next perfect square will be $k=18$ and will take more than $60+(19^2-18^2)=97$ as the value of $m$ , and hence all other perfect squares will take more than $97$ , and we have $97+k$ as our sum which is greater than $79$ since $k\ge 2$ .

its viperstrike. theres a third way. this problem is very tricky!!!

We must ignore the cases where $a_n=1$ , as required by the problem statement. Note as in the previous solution that So write $a_n=\sqrt{400-k}$ for some integers $k,a_n$ . We want to minimize $a_n+n$ . Note if $a_n$ is even ( $k$ is even) then $n=1.5k+2$ , and if $a_n$ is odd ( $k$ is odd), $n=1.5k+1.5$ . Write $k=400-a_n^2$ and note that $a_n$ is at most $19$ . In the case that $a_n$ is odd, we have $a_n+n=a_n+1.5(400-a_n^2)+1.5$ . This is a quadratic in $a_n$ and is strictly decreasing for $a_n>0$ , so to minimize we should choose $a_n=19$ , which gives $a_n+n=79$ . Do the case for $a_n$ even as well, you find that $79$ is the lowest.

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Mock AIME II 2012 Problems/Problem 2

Problem

Solution