- $b_1 > b_2 > \cdots > b_{a_n}$,
- for each $1 \leq i \leq a_n, b_i$ is a power of three, and
- $b_1 = 3^{2021}$
Problem - 01 - National Math Camp 2021 Mock Exam - "Not as bad as it looks"
- Anindya Biswas
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Let $a_1 \leq a_2 \leq a_3 \cdots \leq a_n$ be a sequence of positive integers. For $1 \leq i \leq a_n$, let $b_i$ be the number of terms in the sequence that are not smaller than $i$. It is given that,
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Mehrab4226
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Re: Problem - 01 - National Math Camp 2021 Mock Exam - "Not as bad as it looks"
I am not getting, why this question should have a single answer??? If it has multiple answers, then how should I express those answers???Anindya Biswas wrote: ↑Thu May 13, 2021 12:03 amLet $a_1 \leq a_2 \leq a_3 \cdots \leq a_n$ be a sequence of positive integers. For $1 \leq i \leq a_n$, let $b_i$ be the number of terms in the sequence that are not smaller than $i$. It is given that,Find the sum of $\sum_{j=1}^na_j$ over all possible sequences $\{a_i\}$.
- $b_1 > b_2 > \cdots > b_{a_n}$,
- for each $1 \leq i \leq a_n, b_i$ is a power of three, and
- $b_1 = 3^{2021}$
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
-Henri Poincaré
- Anindya Biswas
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- Joined:Fri Oct 02, 2020 8:51 pm
- Location:Magura, Bangladesh
- Contact:
Re: Problem - 01 - National Math Camp 2021 Mock Exam - "Not as bad as it looks"
Yeah it also confused me at first. It is asking for the sum of all possible values of $\sum_{j=1}^na_j$. That is, sum of the sum!Mehrab4226 wrote: ↑Thu May 13, 2021 1:18 pmI am not getting, why this question should have a single answer??? If it has multiple answers, then how should I express those answers???Anindya Biswas wrote: ↑Thu May 13, 2021 12:03 amLet $a_1 \leq a_2 \leq a_3 \cdots \leq a_n$ be a sequence of positive integers. For $1 \leq i \leq a_n$, let $b_i$ be the number of terms in the sequence that are not smaller than $i$. It is given that,Find the sum of $\sum_{j=1}^na_j$ over all possible sequences $\{a_i\}$.
- $b_1 > b_2 > \cdots > b_{a_n}$,
- for each $1 \leq i \leq a_n, b_i$ is a power of three, and
- $b_1 = 3^{2021}$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: Problem - 01 - National Math Camp 2021 Mock Exam - "Not as bad as it looks"
I am not sure the answer is ok or not as the answer looks quite arbitrary.Anindya Biswas wrote: ↑Thu May 13, 2021 12:03 amLet $a_1 \leq a_2 \leq a_3 \cdots \leq a_n$ be a sequence of positive integers. For $1 \leq i \leq a_n$, let $b_i$ be the number of terms in the sequence that are not smaller than $i$. It is given that,Find the sum of $\sum_{j=1}^na_j$ over all possible sequences $\{a_i\}$.
- $b_1 > b_2 > \cdots > b_{a_n}$,
- for each $1 \leq i \leq a_n, b_i$ is a power of three, and
- $b_1 = 3^{2021}$
Last edited by Mehrab4226 on Thu May 13, 2021 7:45 pm, edited 1 time 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
-Henri Poincaré
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: Problem - 01 - National Math Camp 2021 Mock Exam - "Not as bad as it looks"
Too wordy, I know. Better solution probably existsMehrab4226 wrote: ↑Thu May 13, 2021 5:21 pmI am not sure the answer is ok or not as the answer looks quite arbitrary.Anindya Biswas wrote: ↑Thu May 13, 2021 12:03 amLet $a_1 \leq a_2 \leq a_3 \cdots \leq a_n$ be a sequence of positive integers. For $1 \leq i \leq a_n$, let $b_i$ be the number of terms in the sequence that are not smaller than $i$. It is given that,Find the sum of $\sum_{j=1}^na_j$ over all possible sequences $\{a_i\}$.
- $b_1 > b_2 > \cdots > b_{a_n}$,
- for each $1 \leq i \leq a_n, b_i$ is a power of three, and
- $b_1 = 3^{2021}$
The solution may be a lot difficult to understand. But if any part is not understandable I will appreciate it if the reader asks about it. And the solution can also be wrong.
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
-Henri Poincaré