Key Concepts: Combinatorial Construction and Counting

Hello little mathematicians! Today, we are going to solve a fun math problem together. Imagine you are playing a game with your friend. You have to choose six different natural numbers and arrange them in a circle. Your friend will then pick a number, and you will decide whether to go clockwise or counter-clockwise, labeling the numbers 1 to 6 in order. The tricky part is, each number must be divisible by its label. What is the smallest possible sum of the six numbers you can choose?

Let’s use letters to represent the numbers and break down this problem step by step.

    a
f       b
e       c
    d

First, the number labeled 1 can be any number because any number can be divided by 1. Easy peasy!

Next, let’s look at the number labeled 4. No matter which direction you go, the position of the number 4 is always the same. So, all of the six numbers must be divisible by 4. This also means that the number labeled 2 (which is half of 4) is also taken care of. Two birds with one stone!

Now, we only need to worry about the numbers labeled 3, 5, and 6. Since all the numbers are divisible by 4, if a number is divisible by 3, it will also be divisible by 6. So, we need to make sure that the numbers labeled 3 and 6 are divisible by 3, and the number labeled 5 is divisible by 5.

If $a$ is the starting point, then in the clockwise direction, $c$ is number 3 and $f$ is number 6. In the counter-clockwise direction, $e$ is number 3 and $b$ is number 6. So, either $c$ and $f$ or $b$ and $e$ must be divisible by 3.

By the same logic, if $b$ or $e$ is the starting point, either $a$ and $d$ or $c$ and $f$ must be divisible by 3. If $c$ or $f$ is the starting point, either $a$ and $d$ or $b$ and $e$ must be divisible by 3.

So, we need at least two pairs of numbers to be divisible by 3. Let’s say $b$, $e$, $c$, and $f$ are all divisible by 3.

Now, let’s look at the number 5. If $a$ is the starting point, then either $e$ or $c$ must be divisible by 5. Similarly, if $b$ is the starting point, either $d$ or $f$ must be divisible by 5. If $c$ is the starting point, either $e$ or $a$ must be divisible by 5. If $d$ is the starting point, either $b$ or $f$ must be divisible by 5. If $e$ is the starting point, either $a$ or $c$ must be divisible by 5. If $f$ is the starting point, either $b$ or $d$ must be divisible by 5.

So, we need at least four numbers to be divisible by 5. Let’s say $a$, $b$, $e$, and $f$ are all divisible by 5.

In conclusion, all six numbers must be multiples of 4, four of them must be multiples of 3, and four of them must be multiples of 5. The smallest possible numbers that meet these conditions are 12, 24, 20, 40, 60, and 120. Their sum is 276, which is the smallest possible sum of the six numbers.

I hope you enjoyed solving this problem as much as I did. Remember, math is not about memorizing formulas, but about thinking creatively and logically. Keep practicing, and you will become a math whiz in no time!