## BdMO National 2021 Higher Secondary Problem 12

Anindya Biswas
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Joined:Fri Oct 02, 2020 8:51 pm
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BdMO National 2021 Higher Secondary Problem 12
একটা ফাংশন $$g:\mathbb{Z}\to\mathbb{Z}$$-কে বিশেষণ বলা হবে যদি যেকোনো দুটো পূর্ণসংখ্যা $$m$$ আর $$n$$-এর জন্য $$g(m)+g(n)>\max(m^2, n^2)$$ হয়। $$f$$ এমন একটা বিশেষণ ফাংশন যেন $$f(1)+f(2)+\cdots +f(30)$$-এর মান সর্বনিম্ন হয়। $$f(25)$$-এর সর্বনিম্ন সম্ভাব্য মান বের করো।

A function $g:\mathbb{Z}\to\mathbb{Z}$ is called adjective if $g(m)+g(n)>\max{(m^2,n^2)}$ for any pair of integers $m$ and $n$. Let $f$ be an adjective function such that the value of $f(1)+f(2)+f(3)+\cdots+f(30)$ is minimized. Find the smallest possible value of $f(25)$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

F Nishat
Posts:7
Joined:Fri Apr 16, 2021 4:21 pm

### Re: BdMO National 2021 Higher Secondary Problem 12

Let $S(g)=g(1)+g(2)+\cdots+g(30)=\{g(1)+g(30)\}+\{g(2)+g(29)\}+\cdots+\{g(15)+g(16)\}\ge(30^2+1)+(29^2+1)+(28^2+1)+\cdots(16^2+1)=\sum_{i=16}^{30}\{i^2+1\}.$
So, the minimized value of $S(g)$ is $m=\sum_{i=16}^{30}\{i^2+1\}=f(1)+f(2)+\cdots+f(30).$
Let $j$ be some positive integers such that $j\in[1,15]$. Now, note that $g(j)+g(i)-i^2\ge{1}$ and we achieve $m$ when $g(j)+g(i)-i^2=1$ for all values of $j$ and $i$. This is possible only when $g(1)=g(2)=\cdots=g(15)=k$, which yields $g(i)=i^2-k+1$.
Now, we require the followings:
$\bullet$ $g(1)+g(15)\ge{15^2+1}\Rightarrow{k\ge{113}}$,
$\bullet$ $g(16)\ge{\lceil{\frac{16^2+1}{2}}\rceil}\Rightarrow{k\le{128}}$.
So, $113\le{k}\le{128}$ and we have the function $f$ defined such that:
$f_k(n)=\begin{cases}k& n\in[1,15] \\ n^2-k+1& n\in[16,30] \\ \end{cases}$
We have the smallest value of $f(25)$, when $k=128$. Thus, $f(25)=25^2-128+1=\boxed{498}$.
*Edited after remark of @below.*
Last edited by F Nishat on Mon Apr 26, 2021 12:20 am, edited 1 time in total.
"But in my opinion, all things in nature occur mathematically."-Rene Descartes

Anindya Biswas
Posts:263
Joined:Fri Oct 02, 2020 8:51 pm
Contact:

### Re: BdMO National 2021 Higher Secondary Problem 12

F Nishat wrote:
Sun Apr 25, 2021 4:13 pm
Let $S(g)=g(1)+g(2)+\cdots+g(30)=\{g(1)+g(30)\}+\{g(2)+g(29)\}+\cdots+\{g(15)+g(16)\}\ge(30^2+1)+(29^2+1)+(28^2+1)+\cdots(16^2+1)=\sum_{i=16}^{30}\{i^2+1\}.$
So, the minimized value of $S(g)$ is $m=\sum_{i=16}^{30}\{i^2+1\}=f(1)+f(2)+\cdots+f(30).$
Let $j$ be some positive integers such that $j\in[1,15]$. Now, note that $g(j)+g(i)-i^2\ge{1}$ and we achieve $m$ when $g(j)+g(i)-i^2=1$ for all values of $j$ and $i$. This is possible only when $g(1)=g(2)=\cdots=g(15)=k$, which yields $g(i)=i^2-k+1$.
Now, we require the followings:
$\bullet$ $g(1)+g(15)\ge{15^2+1}\Rightarrow{k\ge{113}}$,
$\bullet$ $g(16)\ge{1}\Rightarrow{k\le{256}}$.
So, $113\le{k}\le{256}$ and we have the function $f$ defined such that:
$f_k(n)=\begin{cases}k& n\in[1,15] \\ n^2-k+1& n\in[16,30] \\ \end{cases}$
We have the smallest value of $f(25)$, when $k=256$. Thus, $f(25)=25^2-256+1=\boxed{370}$.
$k=256$ is not achievable. To demonstrate this, note that
$g(16)+g(16)\geq16^2+1$
$\Rightarrow g(16)\geq 129$
$\Rightarrow 16^2-k+1\geq129$
$\Rightarrow k\leq128$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

F Nishat
Posts:7
Joined:Fri Apr 16, 2021 4:21 pm

### Re: BdMO National 2021 Higher Secondary Problem 12

Anindya Biswas wrote:
Sun Apr 25, 2021 11:44 pm
F Nishat wrote:
Sun Apr 25, 2021 4:13 pm
Let $S(g)=g(1)+g(2)+\cdots+g(30)=\{g(1)+g(30)\}+\{g(2)+g(29)\}+\cdots+\{g(15)+g(16)\}\ge(30^2+1)+(29^2+1)+(28^2+1)+\cdots(16^2+1)=\sum_{i=16}^{30}\{i^2+1\}.$
So, the minimized value of $S(g)$ is $m=\sum_{i=16}^{30}\{i^2+1\}=f(1)+f(2)+\cdots+f(30).$
Let $j$ be some positive integers such that $j\in[1,15]$. Now, note that $g(j)+g(i)-i^2\ge{1}$ and we achieve $m$ when $g(j)+g(i)-i^2=1$ for all values of $j$ and $i$. This is possible only when $g(1)=g(2)=\cdots=g(15)=k$, which yields $g(i)=i^2-k+1$.
Now, we require the followings:
$\bullet$ $g(1)+g(15)\ge{15^2+1}\Rightarrow{k\ge{113}}$,
$\bullet$ $g(16)\ge{1}\Rightarrow{k\le{256}}$.
So, $113\le{k}\le{256}$ and we have the function $f$ defined such that:
$f_k(n)=\begin{cases}k& n\in[1,15] \\ n^2-k+1& n\in[16,30] \\ \end{cases}$
We have the smallest value of $f(25)$, when $k=256$. Thus, $f(25)=25^2-256+1=\boxed{370}$.
$k=256$ is not achievable. To demonstrate this, note that
$g(16)+g(16)\geq16^2+1$
$\Rightarrow g(16)\geq 129$
$\Rightarrow 16^2-k+1\geq129$
$\Rightarrow k\leq128$
You are right. Note that doing the same thing as you did for $g(j)$ (where $j>16$), we get maximum value $k_{max}>{128}$. So, the maximum value of $k$ is indeed $128$.
So, I edited the solution. And thanks for pointing out the mistake . And also sorry for making this silly mistake .
Last edited by F Nishat on Mon Apr 26, 2021 4:02 am, edited 1 time in total.
"But in my opinion, all things in nature occur mathematically."-Rene Descartes

Anindya Biswas
Posts:263
Joined:Fri Oct 02, 2020 8:51 pm
Contact:

### Re: BdMO National 2021 Higher Secondary Problem 12

F Nishat wrote:
Sun Apr 25, 2021 4:13 pm
Let $j$ be some positive integers such that $j\in[1,15]$. Now, note that $g(j)+g(i)-i^2\ge{1}$ and we achieve $m$ when $g(j)+g(i)-i^2=1$ for all values of $j$ and $i$. This is possible only when $g(1)=g(2)=\cdots=g(15)=k$, which yields $g(i)=i^2-k+1$.
I understood the part that $g(j)+g(i)=i^2+1$ when $(i,j)\in\{(30,1),(29,2),(28,3),\dots,(16,15)\}$. But why this must be true for all $i$ and $j$?
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

F Nishat
Posts:7
Joined:Fri Apr 16, 2021 4:21 pm

### Re: BdMO National 2021 Higher Secondary Problem 12

Anindya Biswas wrote:
Mon Apr 26, 2021 12:29 am
F Nishat wrote:
Sun Apr 25, 2021 4:13 pm
Let $j$ be some positive integers such that $j\in[1,15]$. Now, note that $g(j)+g(i)-i^2\ge{1}$ and we achieve $m$ when $g(j)+g(i)-i^2=1$ for all values of $j$ and $i$. This is possible only when $g(1)=g(2)=\cdots=g(15)=k$, which yields $g(i)=i^2-k+1$.
I understood the part that $g(j)+g(i)=i^2+1$ when $(i,j)\in\{(30,1),(29,2),(28,3),\dots,(16,15)\}$. But why this must be true for all $i$ and $j$?
Notice that their sum must be, $m=\sum_{i=16}^{30}(i^2+1)$. Now, let $(j_1,i_1)$ be a pair such that $g(j_1)+g(i_1)=i_1^2+a+1$ ($a$ is a positive integer) and other pairs be such that $g(j)+g(i)$ is at least $i^2+1$. Then, there exists at least one pair $(j_2,i_2)$ such that $g(j_2)+g(i_2)$ is at least $i_2^2-a+1$, which we can't have. So, $g(j)+g(i)$ must be $i^2+1$, for all $j$ and $i$.
"But in my opinion, all things in nature occur mathematically."-Rene Descartes