Subfield: Number Theory


Question: According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulous for a positive integer $m$, then they are all fantabulous. Is the number $2021^{2021}$ fantabulous?

Solution:
Consider the sequence of positive integers $m, 3 m, 6 m+1,12 m+3,4 m+1,2 m$. Since each number in the sequence is fantabulous if and only if the next one is, we deduce that $m$ is fantabulous if and only if $2 m$ is fantabulous.

Combined with the fact that $m$ is fantabulous if and only if $2 m+1$ is fantabulous, this implies that

$m>1$ is fantabulous if and only if $f(m)=\left[\frac{m}{2}\right]$ is fantabulous. We can apply $f$ sufficiently many times to any positive integer $n$ to conclude that $n$ is fantabulous if and only if 1 is fantabulous. Therefore, the fact that 2021 is fantabulous implies that 1 is fantabulous, which in turn implies that $2021^{2021}$ is fantabulous.
Let $m>1$ be a fantabulous number. Note that at least one of the following four cases must hold.

- Case 1. The number $m$ is odd;

We have $m=2 a+1$ for some positive integer $a$, so $a<m$ is also fantabulous.

- Case 2. The number $m$ is a multiple of 3;

We have $m=3 a$ for some positive integer $a$, so $a<m$ is also fantabulous.

- Case 3. The number $m$ is 4 modulo 6;

We have $m=6 a-2$ for some positive integer $a$. We have the sequence of fantabulous numbers

$$
(6 a-2) \rightarrow(12 a-3) \rightarrow(4 a-1)
$$

so $4 a-1<m$ is also fantabulous.


- Case 4. The number $m$ is 2 modulo 6;

We have $m=6 a+2$ for some positive integer $a$. We have the sequence of fantabulous numbers

$$
(6 a+2) \rightarrow(12 a+5) \rightarrow(36 a+15) \rightarrow(18 a+7) \rightarrow(9 a+3) \rightarrow(3 a+1)
$$

so $3 a+1<m$ is also fantabulous.

In all cases, we see that there is another fantabulous number less than $m$. Since 2021 is fantabulous, it follows that 1 is fantabulous.

Observe that a number $m$ is not fantabulous if and only if all of the elements of the set $\{m, 2 m+1,3 m\}$ are not fantabulous. So, the argument above shows that if there exists a positive integer that is not fantabulous, then 1 would not be fantabulous either. This is a contradiction, so all positive integers are fantabulous and, in particular, $2021^{2021}$ is fantabulous.
The following transformations show that $a$ is fantabulous if and only if $3 a, 3 a+1$ or $3 a+2$ are fantabulous.

$$
\begin{aligned}
& a \rightarrow 3 a \\
& a \rightarrow 2 a+1 \rightarrow 6 a+3 \rightarrow 3 a+1 \\
& a \rightarrow 2 a+1 \rightarrow 4 a+3 \rightarrow 12 a+9 \rightarrow 36 a+27 \rightarrow 18 a+13 \rightarrow 9 a+6 \rightarrow 3 a+2
\end{aligned}
$$

This implies that $a \geq 3$ is fantabulous if and only if $f(a)=\left[\frac{a}{3}\right]$ is fantabulous. We can use this to deduce that 1 and 2 are fantabulous from the fact that 2021 is fantabulous in the following way:

$$
2021 \rightarrow 673 \rightarrow 224 \rightarrow 74 \rightarrow 24 \rightarrow 8 \rightarrow 2 \rightarrow 5 \rightarrow 1
$$

We can apply $f$ sufficiently many times to any positive integer $n$ to arrive at the number 1 or 2 . It follows that every positive integer is fantabulous, so $2021^{2021}$ is fantabulous.