BdMO National 2021 Secondary Problem 9
BdMO National 2021 Secondary Problem 9
Cynthia loves Pokemon and she wants to catch them all. In Victory Road, there are a total of $80$ Pokemon. Cynthia wants to catch as many of them as possible. However, she cannot catch any two Pokemon that are enemies with each other. After exploring around for a while, she makes the following two observations:
1. Every Pokemon in Victory Road is enemies with exactly two other Pokemon.
2. Due to her inability to catch Pokemon that are enemies with one another, the maximum number of Pokemon she can catch is equal to $n$.
What is the sum of all possible values of $n$?
1. Every Pokemon in Victory Road is enemies with exactly two other Pokemon.
2. Due to her inability to catch Pokemon that are enemies with one another, the maximum number of Pokemon she can catch is equal to $n$.
What is the sum of all possible values of $n$?
"When you change the way you look at things, the things you look at change."  Max Planck
 Swapnil Barua
 Posts: 14
 Joined: Tue Apr 13, 2021 2:55 pm
Re: BdMO National 2021 Secondary Problem 9
Choose 1 to 10 numbers
1,2,3 and 4,5,6 and 7,8,9 and 10 (1,2,3 and 4,5,6 and 7,8,9 every 3 of them are enemies with each other)
1 isn't enemy with 4, 2 isn't enemy with 5, 3 isn't enemy with 6, (all 6 here)
There are 7,8,9 and 10 left
From them you can take any 1 number with 10 (all 2 here)
2+6=8
Cynthia can take From 10 pokemons 8 (which aren't enemy)
From 110 pokemons 110×8÷10=88
1,2,3 and 4,5,6 and 7,8,9 and 10 (1,2,3 and 4,5,6 and 7,8,9 every 3 of them are enemies with each other)
1 isn't enemy with 4, 2 isn't enemy with 5, 3 isn't enemy with 6, (all 6 here)
There are 7,8,9 and 10 left
From them you can take any 1 number with 10 (all 2 here)
2+6=8
Cynthia can take From 10 pokemons 8 (which aren't enemy)
From 110 pokemons 110×8÷10=88
Re: BdMO National 2021 Secondary Problem 9
You're misunderstanding the problem. The number of maximum pokemon is entirely dependent on how they are set as enemies ie. the graph. So, n can change depending on how the graph looks. And thus creating multiple values of $n$. Also, I don't quite understand how you're taking $8$ from $10$.Swapnil Barua wrote: ↑Tue Apr 13, 2021 3:03 pmChoose 1 to 10 numbers
1,2,3 and 4,5,6 and 7,8,9 and 10 (1,2,3 and 4,5,6 and 7,8,9 every 3 of them are enemies with each other)
1 isn't enemy with 4, 2 isn't enemy with 5, 3 isn't enemy with 6, (all 6 here)
There are 7,8,9 and 10 left
From them you can take any 1 number with 10 (all 2 here)
2+6=8
Cynthia can take From 10 pokemons 8 (which aren't enemy)
From 110 pokemons 110×8÷10=88
"When you change the way you look at things, the things you look at change."  Max Planck
 Swapnil Barua
 Posts: 14
 Joined: Tue Apr 13, 2021 2:55 pm
Re: BdMO National 2021 Secondary Problem 9
7,8,9 are enemies with each other
You can't take any of them
You must take 7,10 or 8,10 or 9,10( You can choose one pair)
I didn't take any multiple value
In the last there will be 2 pokemons left from 110
If you take 7,10 the remaining pokemons will be 8,9
and if you take 8,10 the remaining pokemons will be 7,9
same goes for 9,10( remaining 7,8)
You can't take 7,8 or,8,9 or 7,9 (enemies with each other)
You can't take any of them
You must take 7,10 or 8,10 or 9,10( You can choose one pair)
I didn't take any multiple value
In the last there will be 2 pokemons left from 110
If you take 7,10 the remaining pokemons will be 8,9
and if you take 8,10 the remaining pokemons will be 7,9
same goes for 9,10( remaining 7,8)
You can't take 7,8 or,8,9 or 7,9 (enemies with each other)
 Mehrab4226
 Posts: 208
 Joined: Sat Jan 11, 2020 1:38 pm
 Location: Dhaka, Bangladesh
Re: BdMO National 2021 Secondary Problem 9
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

 Posts: 170
 Joined: Sat Jan 02, 2021 9:28 pm
Re: BdMO National 2021 Secondary Problem 9
she can catch any integer number of pokemon within $[27,40]$. She can catch those numbers if we take a few of the vertices of $cycle3$ and add them up in $cycle5$ or the cycle with the maximum vertices. Elaborate pls
Hmm..Hammer...Treat everything as nail
 Mehrab4226
 Posts: 208
 Joined: Sat Jan 11, 2020 1:38 pm
 Location: Dhaka, Bangladesh
Re: BdMO National 2021 Secondary Problem 9
At first we can agree,Asif Hossain wrote: ↑Thu Apr 22, 2021 2:32 pmshe can catch any integer number of pokemon within $[27,40]$. She can catch those numbers if we take a few of the vertices of $cycle3$ and add them up in $cycle5$ or the cycle with the maximum vertices. Elaborate pls
$\lf \frac{3}{2}\rf +\cdots 25 times \cdots \lf \frac{3}{2} \rf + \lf \frac{5}{2} \rf =27$
Now this is the condition when there is $25$ cycle3(cycle with 3 vertices) and $1$ cycle5(Cycle with 5 vertices). Now if we take a 3 vertices from a single cycle3 and add them in the cycle5 this will make a cycle8. So we have $24$ cycle3 and $1$ cycle8.
So the total number of pokemon Cynthia can catch is $1\times 24+4=28$
We can do similar steps again and get,
$23$ cycle3 and $1$ cycle11, which makes the number of pokemon $23+5=28$
Do the same again and we'll get $22+7=29$.
In this way, we can get all the numbers between 27 and 40 inclusive.
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