2018 BDMO NT exam P4

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Ananya Promi
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2018 BDMO NT exam P4

Unread post by Ananya Promi » Thu Apr 12, 2018 3:43 pm

Let $n$ be an even positive integer. A sequence of $n$ real numbers is called complete if for every integer $m$ with $$1 \leq m \leq n$$ either the sum of the first $m$ terms or the sum of the last $m$ terms is integral. Determine the minimum number of the integers in a complete sequence of $n$ numbers.

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Atonu Roy Chowdhury
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Re: 2018 BDMO NT exam P4

Unread post by Atonu Roy Chowdhury » Sat Apr 14, 2018 8:21 pm

শুভ নববর্ষ ১৪২৫
We will show that there are at least $2$ integers in the sequence. Assume the contrary, i.e. assume there are at most $1$ integer.
Let $\{a_i\}_{i=1}^{n}$ be the sequence.
For $m=1$, we get either $a_1$ or $a_n$ is integer. WLOG, $a_1$ is integer.
For $m=2$, either $a_1+a_2$ or $a_{n-1}+a_n$ is integer. If $a_1+a_2$ is integer, then $a_2$ must be an integer which contradicts our assumption. So, $a_{n-1}+a_n \in \mathbb{Z}$
Similarly for $m=3$, we get $a_2+a_3 \in \mathbb{Z} $
Analogously, we get some pairwise sums which are integers.
For $m=n$, $a_1 + a_2+ a_3 +\cdots + a_{n-1} + a_n \in \mathbb{Z} \Rightarrow a_2+ a_3 +\cdots + a_{n-1} + a_n \in \mathbb{Z}$
$n-1$ is odd. So we can't divide the numbers $a_2, a_3, \cdots, a_{n-1}, a_n$ into some pairs.
So, $a_1 + a_2+ a_3 +\cdots + a_{n-1} + a_n$ can't be an integer if there are at most $1$ integer.
Contradiction!

Now, all we need to do is to find such a construction where there are $2$ integers and all our conditions are satisfied.
When, $n=4k$ our construction will be \[\underbrace {1, .5, .5, \cdots .5, 1}_{first 2k terms} , \underbrace {.5 .5, .5, \cdots .5, .5}_{second 2k terms} \]
When, $n=4k+2$ our construction will be \[\underbrace {1, .5, .5, \cdots .5, .5}_{first 2k terms} , \underbrace {1 .5, .5, \cdots .5, .5}_{second 2k terms} \]
So, the answer is $2$
This was freedom. Losing all hope was freedom.

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samiul_samin
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Re: 2018 BDMO NT exam P4

Unread post by samiul_samin » Mon Feb 18, 2019 11:51 pm

Ananya Promi wrote:
Thu Apr 12, 2018 3:43 pm
Let $n$ be an even positive integer. A sequence of $n$ real numbers is called complete if for every integer $m$ with $$1 \leq m \leq n$$ either the sum of the first $m$ terms or the sum of the last $m$ terms is integral. Determine the minimum number of the integers in a complete sequence of $n$ numbers.
Is this problem taken from National Math Camp question?

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