Hello young mathematicians! Today, we’re going to explore a fun and interesting problem from the world of elementary math Olympiad. This problem involves a special sequence of numbers, and we’re going to solve it together. So, let’s dive in!

Imagine we have a sequence of numbers, let’s call them $a_1, a_2, a_3, …, a_n$. This isn’t just any sequence, though. It has two very special rules:

  1. If you pick any two numbers that are five places apart, the one in front is always bigger. In other words, $a_6 > a_1$.
  2. If you pick any two numbers that are four places apart, the one in the back is always bigger. So, $a_5 < a_1$.

Now, the question is: what’s the maximum number of terms this sequence can have?

Let’s try to figure this out together.

First, let’s write down the implications of our rules:

  • $a_6 > a_1$
  • $a_1 > a_7 > a_2$
  • $a_2 > a_8 > a_3$
  • $a_3 > a_9 > a_4$
  • $a_4 > a_{10} > a_5$
  • $a_5 > a_{11} > a_6$
  • …

Do you see what’s happening here? If we look at these inequalities as a whole, we see a sequence of numbers that is decreasing, but there’s a problem. $a_6$ is appearing twice, and it’s both smaller and larger than $a_{11}$. That’s a contradiction!

So, we can conclude that our sequence cannot have an $a_{11}$. Therefore, the sequence can have at most 10 terms.

Let’s check if a sequence with 10 terms can satisfy our rules. Let’s take the numbers from 1 to 10 and arrange them in descending order. Then, we distribute them to the terms $a_1$ to $a_{10}$ like this:

$a_6, a_1, a_7, a_2, a_8, a_3, a_9, a_4, a_{10}, a_5 = 10, 9, 8, 7, 6, 5, 4, 3, 2, 1$

So, our sequence becomes: $a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10} = 9, 7, 5, 3, 1, 10, 8, 6, 4, 2$

And voila! We have our magic sequence of 10 terms that follows our special rules.

I hope you enjoyed this journey into the world of math Olympiad problems. Remember, math is not just about numbers, it’s also about patterns and logic. So, keep exploring and happy problem-solving!