BdMO National 2021 Secondary P5

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
User avatar
Pro_GRMR
Posts:46
Joined:Wed Feb 03, 2021 1:58 pm
BdMO National 2021 Secondary P5

Unread post by Pro_GRMR » Sat Apr 10, 2021 7:07 pm

\(g(x):\mathbb{Z}\to\mathbb{Z}\) একটা ফাংশন যেটা নিচের শর্তকে মেনে চলে।
\[g(x)+g(y)=g(x+y)-xy\]
যদি f \(g(23)=0\) হয়, তাহলে \(g(40)\)-এর সম্ভাব্য সব মানের যোগফল কত?

$g(x) : \mathbb{Z} \to \mathbb{Z}$ that satisfies

\[f(x+y)-xy =f(x)+f(y)\]
If $f(23)=0$, what is the sum of all possible values of $f(40)$?
Last edited by Pro_GRMR on Sat Apr 10, 2021 7:41 pm, edited 2 times in total.
"When you change the way you look at things, the things you look at change." - Max Planck

User avatar
Pro_GRMR
Posts:46
Joined:Wed Feb 03, 2021 1:58 pm

Re: BdMO National 2021 Secondary P5

Unread post by Pro_GRMR » Sat Apr 10, 2021 7:40 pm

Pro_GRMR wrote:
Sat Apr 10, 2021 7:07 pm
$g(x) : \mathbb{Z} \to \mathbb{Z}$ that satisfies
\[f(x+y)-xy =f(x)+f(y)\]
If $f(23)=0$, what is the sum of all possible values of $f(40)$?
Setting y = 1 and shuffling the terms we get: f(x+1)-f(x)=f(1)+x
Now we telescope the sum:
\begin{align*}
f(x)-f(x-1) &= f(1)+x-1\\
f(x-1)-f(x-2)&=f(1)+x-2\\
\cdots\cdots\cdots\cdots\cdots&=\cdots\cdots\cdots\cdots\cdots\\
f(3)-f(2)&=f(1)+2\\
f(2)-f(1)&=f(1)+1\\
\end{align*}
We get $f(x)=xf(1)+\frac{n(n-1)}{2}$
So setting $x=23$,
$f(23)=23f(1)+\frac{23\cdot22}{2}$
$\Longrightarrow f(1)=-11$
Again after setting $x=40$,
$f(40)=40f(1)+\frac{40\cdot39}{2}$
$\Longrightarrow f(40)=40(-11)+\frac{40\cdot39}{2}$
$\Longrightarrow f(40)=\boxed{340}$
"When you change the way you look at things, the things you look at change." - Max Planck

Post Reply