BdMO Regional 2021 Secondary P2

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Pro_GRMR
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BdMO Regional 2021 Secondary P2

Unread post by Pro_GRMR » Tue Mar 30, 2021 4:07 pm

You have six boxes numbered $1, 2, 3, 4, 5$ and $6$ respectively.Your friend has distributed $n$ balls among these boxes. What is the smallest possible value of $n$ for which you can guarantee that there is at least one box that contains at least as many balls as the square of the number written on it?
Or,
There are seven students in a class and their roll numbers are $1, 2, 3, 4, 5, 6$ and $7$ respectively. They have distributed $k$ taka among themselves where $k$ is a positive integer. What is the smallest possible value of $k$ for which you can guarantee that there exists a student with at least as much taka as the square of their roll number?
"When you change the way you look at things, the things you look at change." - Max Planck

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Pro_GRMR
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Joined:Wed Feb 03, 2021 1:58 pm

Re: BdMO Regional 2021 Secondary P2

Unread post by Pro_GRMR » Tue Mar 30, 2021 4:15 pm

Pro_GRMR wrote:
Tue Mar 30, 2021 4:07 pm
You have six boxes numbered $1, 2, 3, 4, 5$ and $6$ respectively.Your friend has distributed $n$ balls among these boxes. What is the smallest possible value of $n$ for which you can guarantee that there is at least one box that contains at least as many balls as the square of the number written on it?
Or,
There are seven students in a class and their roll numbers are $1, 2, 3, 4, 5, 6$ and $7$ respectively. They have distributed $k$ taka among themselves where $k$ is a positive integer. What is the smallest possible value of $k$ for which you can guarantee that there exists a student with at least as much taka as the square of their roll number?
The problem is solved in a similar way for both varieties.

Let's think about the highest value of $k$ for which no one gets as much taka as the square of their roll number. They can get $0, 3, 8, 15, 24, 35$ and $48$ taka respectively. But if there is $1$ more taka, someone has to get it and they will have as much taka as the square of their roll number.
So the minimum value of such $k$ would be $0+3+8+15+24+35+48+1=\boxed{134}$.
Using the same logic in the first variety as above, the minimum value of such $n$ would be $0+3+8+15+24+35+1=\boxed{86}$.
"When you change the way you look at things, the things you look at change." - Max Planck

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