BdMO National 2021 Higher Secondary Problem 8

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Anindya Biswas
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BdMO National 2021 Higher Secondary Problem 8

Unread post by Anindya Biswas » Sun Apr 11, 2021 9:48 pm

শাকুর আর তিহাম একটা খেলা খেলছে। প্রথমে, শাকুর \(1000\)-এর চেয়ে বড় না এমন একটা ধনাত্মক পূর্ণসংখ্যা বাছাই করে। তারপরে তিহাম তার থেকে ছোট আরেকটা ধনাত্মক পূর্ণসংখ্যা বাছাই করে। তারা এভাবে পালাক্রমে ছোট থেকে ছোটতর পূর্ণসংখ্যা বাছাই করতে থাকে যতক্ষণ পর্যন্ত কেউ \(1\) বাছাই না করে। কেউ \(1\) বাছাই করার পর ওই পর্যন্ত বাছাইকৃত সবগুলো সংখ্যা যোগ করা হয়। যে \(1\) বাছাই করে সে জেতে যদি আর কেবল যদি এই যোগফলটা একটা পূর্ণবর্গ সংখ্যা হয়। যদি যোগফলটা পূর্ণবর্গ না হয়, তাহলে অপরজন জেতে। এমন সম্ভাব্য সব \(n\)-এর যোগফল কত যেন যদি শাকুর \(n\) বলে খেলাটা শুরু করে, তাহলে তার একটা জেতার স্ট্র্যাটেজি আছে?

Shakur and Tiham are playing a game. Initially, Shakur picks a positive integer not greater than $1000$. Then Tiham picks a positive integer strictly smaller than that. Then they keep on doing this taking turns to pick progressively smaller and smaller positive integers until someone picks $1$. After that, all the numbers that have been picked so far are added up. The person picking the number $1$ wins if and only if this sum is a perfect square. Otherwise, the other player wins. What is the sum of all possible values of $n$ such that if Shakur starts with the number $n$, he has a winning strategy?
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mehrab4226
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Re: BdMO National 2021 Higher Secondary Problem 8

Unread post by Mehrab4226 » Sun Apr 11, 2021 10:00 pm

Anindya Biswas wrote:
Sun Apr 11, 2021 9:48 pm
Shakur and Tiham are playing a game. Initially, Shakur picks a positive integer not greater than $1000$. Then Tiham picks a positive integer strictly smaller than that. Then they keep on doing this taking turns to pick progressively smaller and smaller positive integers until someone picks $1$. After that, all the numbers that have been picked so far are added up. The person picking the number $1$ wins if and only if this sum is a perfect square. Otherwise, the other player wins. What is the sum of all possible values of $n$ such that if Shakur starts with the number $n$, he has a winning strategy?
I don't think Shakur can win. Since if he takes $n$ in his first turn, Tiham can easily choose 1 which does not break any rule, and Tiham also has a square number making Tiham the winner. Or I didn't understand the question
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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Anindya Biswas
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Re: BdMO National 2021 Higher Secondary Problem 8

Unread post by Anindya Biswas » Sun Apr 11, 2021 10:08 pm

Mehrab4226 wrote:
Sun Apr 11, 2021 10:00 pm
Anindya Biswas wrote:
Sun Apr 11, 2021 9:48 pm
Shakur and Tiham are playing a game. Initially, Shakur picks a positive integer not greater than $1000$. Then Tiham picks a positive integer strictly smaller than that. Then they keep on doing this taking turns to pick progressively smaller and smaller positive integers until someone picks $1$. After that, all the numbers that have been picked so far are added up. The person picking the number $1$ wins if and only if this sum is a perfect square. Otherwise, the other player wins. What is the sum of all possible values of $n$ such that if Shakur starts with the number $n$, he has a winning strategy?
I don't think Shakur can win. Since if he takes $n$ in his first turn, Tiham can easily choose 1 which does not break any rule, and Tiham also has a square number making Tiham the winner. Or I didn't understand the question
If Tiham takes $1$ and the sum isn't a perfect square, then he will lose. (The sum means the sum of all number chosen by both of them.)
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mehrab4226
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Re: BdMO National 2021 Higher Secondary Problem 8

Unread post by Mehrab4226 » Wed Apr 14, 2021 2:01 am

We claim there are only 2 values of n for which Shakur has a winning strategy.
But first,
Note that if Shakur starts with any number say $n$ and $n+3$ is not a square number then Tiham can choose $2$ and then Shakur has to choose $1$ and lose the game.

Now our Test subjects are narrowed down to $31$ as $32^2-3=1021>1000$ which is totally useless.

Now, let,
Shakur first chooses $n^2-3$, then Tiham can choose $(n+1)^2-n^2$
So the sum of the chosen numbers is $(n+1)^2-3$ so Shakur cannot choose $2$, so he should choose something that will make the sum $k^2-3$ because if it is not then Tiham can choose $2$ and win. To do that the smallest number that Shakur may choose is $(n+2)^2-(n+1)^2$ which will make the sum $(n+2)^2-3$. First why this is the smallest number that Shakur can choose?. Well if he chooses any less number then that then the sum,$S$ will be, $(n+2)^2-3>S>(n+1)^2-3$. So we cannot get our desired sum and Shakur will lose. But Shakur cannot choose $(n+2)^2-(n+1)^2$ either as it is greater than what Tiham last chose. So Shakur will lose.

But there is a small sweet spot. When Tiham cannot choose $(n+1)^2-n^2$.
Tiham cannot choose that if,
$(n+1)^2-n^2>n^2-3$
Or,$2n+1>n^2-3$
Or,$4>n^2-2n$
Or,$4>n(n-2)$
This inequality will hold if $n=1,2,3$
So, the numbers Shakur can start with and have a winning strategy are,
$1^2-3=-2$[This is not acceptable]
$2^2-3=1$
$3^2-3=6$
$2$[Thanks @Aninda Biswas]
So required sum of all the possible values of $n$ is $1+2+6=9$(Ans) idk
Last edited by Mehrab4226 on Wed Apr 14, 2021 1:55 pm, edited 2 times in total.
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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Anindya Biswas
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Re: BdMO National 2021 Higher Secondary Problem 8

Unread post by Anindya Biswas » Wed Apr 14, 2021 10:18 am

Mehrab4226 wrote:
Wed Apr 14, 2021 2:01 am
We claim there are only 2 values of n for which Shakur has a winning strategy.
But first,
Note that if Shakur starts with any number say $n$ and $n+3$ is not a square number then Tiham can choose $2$ and then Shakur has to choose $1$ and lose the game.

Now our Test subjects are narrowed down to $31$ as $32^2-3=1021>1000$ which is totally useless.

Now, let,
Shakur first chooses $n^2-3$, then Tiham can choose $(n+1)^2-n^2$
So the sum of the chosen numbers is $(n+1)^2-3$ so Shakur cannot choose $2$, so he should choose something that will make the sum $k^2-3$ because if it is not then Tiham can choose $2$ and win. To do that the smallest number that Shakur may choose is $(n+2)^2-(n+1)^2$ which will make the sum $(n+2)^2-3$. First why this is the smallest number that Shakur can choose?. Well if he chooses any less number then that then the sum,$S$ will be, $(n+2)^2-3>S>(n+1)^2-3$. So we cannot get our desired sum and Shakur will lose. But Shakur cannot choose $(n+2)^2-(n+1)^2$ either as it is greater than what Tiham last chose. So Shakur will lose.

But there is a small sweet spot. When Tiham cannot choose $(n+1)^2-n^2$.
Tiham cannot choose that if,
$(n+1)^2-n^2>n^2-3$
Or,$2n+1>n^2-3$
Or,$4>n^2-2n$
Or,$4>n(n-2)$
This inequality will hold if $n=1,2,3$
So, the numbers Shakur can start with and have a winning strategy are,
$1^2-3=-2$[This is not acceptable]
$2^2-3=1$
$3^2-3=6$
So required sum of all the possible values of $n$ is $1+6=7$(Ans) idk
Shakur also wins if he chooses $2$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mehrab4226
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Location: Dhaka, Bangladesh

Re: BdMO National 2021 Higher Secondary Problem 8

Unread post by Mehrab4226 » Wed Apr 14, 2021 1:53 pm

Anindya Biswas wrote:
Wed Apr 14, 2021 10:18 am
Mehrab4226 wrote:
Wed Apr 14, 2021 2:01 am
We claim there are only 2 values of n for which Shakur has a winning strategy.
But first,
Note that if Shakur starts with any number say $n$ and $n+3$ is not a square number then Tiham can choose $2$ and then Shakur has to choose $1$ and lose the game.

Now our Test subjects are narrowed down to $31$ as $32^2-3=1021>1000$ which is totally useless.

Now, let,
Shakur first chooses $n^2-3$, then Tiham can choose $(n+1)^2-n^2$
So the sum of the chosen numbers is $(n+1)^2-3$ so Shakur cannot choose $2$, so he should choose something that will make the sum $k^2-3$ because if it is not then Tiham can choose $2$ and win. To do that the smallest number that Shakur may choose is $(n+2)^2-(n+1)^2$ which will make the sum $(n+2)^2-3$. First why this is the smallest number that Shakur can choose?. Well if he chooses any less number then that then the sum,$S$ will be, $(n+2)^2-3>S>(n+1)^2-3$. So we cannot get our desired sum and Shakur will lose. But Shakur cannot choose $(n+2)^2-(n+1)^2$ either as it is greater than what Tiham last chose. So Shakur will lose.

But there is a small sweet spot. When Tiham cannot choose $(n+1)^2-n^2$.
Tiham cannot choose that if,
$(n+1)^2-n^2>n^2-3$
Or,$2n+1>n^2-3$
Or,$4>n^2-2n$
Or,$4>n(n-2)$
This inequality will hold if $n=1,2,3$
So, the numbers Shakur can start with and have a winning strategy are,
$1^2-3=-2$[This is not acceptable]
$2^2-3=1$
$3^2-3=6$
So required sum of all the possible values of $n$ is $1+6=7$(Ans) idk
Shakur also wins if he chooses $2$
Oh yeah almost forgot about that!
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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