EGMO 2021 P1
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- Posts:194
- Joined:Sat Jan 02, 2021 9:28 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
- Mehrab4226
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- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: EGMO 2021 P1
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
-Henri Poincaré
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- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Re: EGMO 2021 P1
Hmm..Hammer...Treat everything as nail
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: EGMO 2021 P1
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
-Henri Poincaré