EGMO 2021 P1

For discussing Olympiad Level Number Theory problems
Asif Hossain
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Joined: Sat Jan 02, 2021 9:28 pm

EGMO 2021 P1

Unread post by Asif Hossain » Thu Apr 15, 2021 12:39 pm

The number $2021$ is fantabulous. For any positive integer $m$, if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous?
Hmm..Hammer...Treat everything as nail

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Mehrab4226
Posts: 208
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Location: Dhaka, Bangladesh

Re: EGMO 2021 P1

Unread post by Mehrab4226 » Thu Apr 15, 2021 10:50 pm

At first note,
$2021 \equiv -1\text{ (Mod 3)}$
$\therefore 2021^{2021} \equiv -1 \text{ (Mod 3)}$
So the number should change into $-1$ mod 3. So there are 3 cases
In $\text{mod 3}$
  1. $m \equiv 0$
    $2m+1 \equiv 1$
    $3m \equiv 0$
  2. $m \equiv 1$
    $2m+1 \equiv 0$
    $3m \equiv 0$
  3. $m \equiv -1$
    $2m+1 \equiv -1$
    $3m \equiv 0$
So now we can simplify our problem to just focus on $\{m,2m+1\}$ where both elements must be $-1$ in mod $3$.
Now we can get a sequence,
$m,2m+1,2(2m+1)+1,\cdots$
If $2021^{2021}$ is a fantabulous number it must be in the sequence where $m=2021$.
But in $\text{Mod 4}$ we get,
$2021 \equiv 1$
$2021^{2021} \equiv 1$
In the sequence,
$m \equiv 1$
$2m+1 \equiv 3 \equiv -1$
$2(2m+1)+1 \equiv -1 $
So excluding the 1st term, all the other terms of the sequence are congruent to $-1 \text{ (Mod 4)}$ But since $2021^{2021} \equiv 1 \text{ (Mod 4)}$. $2021^{2021}$ is not in the sequence. So it is not a fantabulous number.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Asif Hossain
Posts: 169
Joined: Sat Jan 02, 2021 9:28 pm

Re: EGMO 2021 P1

Unread post by Asif Hossain » Fri Apr 16, 2021 11:37 am

Mehrab4226 wrote:
Thu Apr 15, 2021 10:50 pm
At first note,
$2021 \equiv -1\text{ (Mod 3)}$
$\therefore 2021^{2021} \equiv -1 \text{ (Mod 3)}$
So the number should change into $-1$ mod 3. So there are 3 cases
In $\text{mod 3}$
  1. $m \equiv 0$
    $2m+1 \equiv 1$
    $3m \equiv 0$
  2. $m \equiv 1$
    $2m+1 \equiv 0$
    $3m \equiv 0$
  3. $m \equiv -1$
    $2m+1 \equiv -1$
    $3m \equiv 0$
So now we can simplify our problem to just focus on $\{m,2m+1\}$ where both elements must be $-1$ in mod $3$.
Now we can get a sequence,
$m,2m+1,2(2m+1)+1,\cdots$
If $2021^{2021}$ is a fantabulous number it must be in the sequence where $m=2021$.
But in $\text{Mod 4}$ we get,
$2021 \equiv 1$
$2021^{2021} \equiv 1$
In the sequence,
$m \equiv 1$
$2m+1 \equiv 3 \equiv -1$
$2(2m+1)+1 \equiv -1 $
So excluding the 1st term, all the other terms of the sequence are congruent to $-1 \text{ (Mod 4)}$ But since $2021^{2021} \equiv 1 \text{ (Mod 4)}$. $2021^{2021}$ is not in the sequence. So it is not a fantabulous number.
Recheck your proof $2021^{2021}$ IS FANTABULOUS.
HINTS
Every number is fantabulous and collatz conjecture(not necessary but if you know you will catch it quickly).
Hmm..Hammer...Treat everything as nail

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Mehrab4226
Posts: 208
Joined: Sat Jan 11, 2020 1:38 pm
Location: Dhaka, Bangladesh

Re: EGMO 2021 P1

Unread post by Mehrab4226 » Wed Apr 21, 2021 2:25 pm

Mehrab4226 wrote:
Thu Apr 15, 2021 10:50 pm
At first note,
$2021 \equiv -1\text{ (Mod 3)}$
$\therefore 2021^{2021} \equiv -1 \text{ (Mod 3)}$
So the number should change into $-1$ mod 3. So there are 3 cases
In $\text{mod 3}$
  1. $m \equiv 0$
    $2m+1 \equiv 1$
    $3m \equiv 0$
  2. $m \equiv 1$
    $2m+1 \equiv 0$
    $3m \equiv 0$
  3. $m \equiv -1$
    $2m+1 \equiv -1$
    $3m \equiv 0$
So now we can simplify our problem to just focus on $\{m,2m+1\}$ where both elements must be $-1$ in mod $3$.
Now we can get a sequence,
$m,2m+1,2(2m+1)+1,\cdots$
If $2021^{2021}$ is a fantabulous number it must be in the sequence where $m=2021$.
But in $\text{Mod 4}$ we get,
$2021 \equiv 1$
$2021^{2021} \equiv 1$
In the sequence,
$m \equiv 1$
$2m+1 \equiv 3 \equiv -1$
$2(2m+1)+1 \equiv -1 $
So excluding the 1st term, all the other terms of the sequence are congruent to $-1 \text{ (Mod 4)}$ But since $2021^{2021} \equiv 1 \text{ (Mod 4)}$. $2021^{2021}$ is not in the sequence. So it is not a fantabulous number.
Ok. There is a problem with the solution,
Note,
$m \to 3m \to 6m+1 \to 12m+3 \to 4m+1 \to 2m $
So if 2m is fantabulous so is m, and it is the same if 2m+1 is fantabulous.
Now,
$2021 \to 1010 \to 505 \to 252 \to 126 \to 63 \to 31 \to 15 \to 7 \to 3 \to 2 \to 1.$
So 1 is fantabulous.
If any number $k$ is not fantabulous, then we can decrease it to 1 and get 1 is not fantabulous which is not true. Contradiction. So all numbers are fantabulous.
$\therefore 2021^{2021}$ is fantabulous.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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