## BdMO National 2021 Higher Secondary Problem 1

Anindya Biswas
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BdMO National 2021 Higher Secondary Problem 1
কোনো ধনাত্মক পূর্ণসংখ্যা $$n$$-এর জন্য $$A(n)$$ হলো $$n$$-কে $$11$$ দিয়ে ভাগ করলে যে ভাগশেষ হয়, সেটা। আর $$T(n)=A(1)+A(2)+A(3)+\cdots + A(n)$$। $$A(T(2021))$$-এর মান বের করো।

For a positive integer $n$, let $A(n)$ be equal to the remainder when $n$ is divided by $11$ and let $T(n)=A(1)+A(2)+A(3)+\dots+A(n)$. Find the value of $A(T(2021))$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Mehrab4226
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### Re: BdMO National 2021 Higher Secondary Problem 1

Anindya Biswas wrote:
Sun Apr 11, 2021 8:54 pm
For a positive integer $n$, let $A(n)$ be equal to the remainder when $n$ is divided by $11$ and let $T(n)=A(1)+A(2)+A(3)+\dots+A(n)$. Find the value of $A(T(2021))$.
$A(1)+A(2)+\cdots +A(11)=1+2+\cdots 10+0=55$
This will repeat for every multiple of 11, like,
$A(12)+A(13)+\cdots +A(22)=55$
Now, $2013=183\times 11$
So,
$A(1)+A(2)+ \cdots +A(2013)+A(2014)+\cdots +A(2021)=55 \times 183 +1+2+ \cdots 8=10101$
Therefore $A(10101)=3$(ans) Idk
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

Pro_GRMR
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### Re: BdMO National 2021 Higher Secondary Problem 1

Anindya Biswas wrote:
Sun Apr 11, 2021 8:54 pm
For a positive integer $n$, let $A(n)$ be equal to the remainder when $n$ is divided by $11$ and let $T(n)=A(1)+A(2)+A(3)+\dots+A(n)$. Find the value of $A(T(2021))$.
Note: All the equivalent relations will be modulo 11.
We notice $A(n) \equiv n$ and thus $T(n) \equiv \frac{n(n+1)}{2}$
So, $T(2021) \equiv \frac{2021(2022)}{2} \equiv 2021 \cdot 1011 \equiv 3 \cdot 1 \equiv 3$ Which is the residue/remainder when $T(2021)$ is divided by $11$.
And so $\boxed{3}$ is the answer.
"When you change the way you look at things, the things you look at change." - Max Planck