complete integral values

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Source: Netherlands TST for BxMO 2017 problem 1

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#1 h Feb 1, 2018, 3:07 PM • 3 Y

Y by Muradjl, Adventure10, PikaPika999

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 of the sum or the sum of the last $m$ terms is integral. Determine
the minimum number of integers in a complete sequence of $n$ numbers.

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#2 h Feb 1, 2018, 7:15 PM • 3 Y

Y by Adventure10, Mango247, PikaPika999

We show that $2$ works, and then we prove that it is minimum.

Define the sequence as follows:
Fill the first position with any integer.
Then, fill the two rightmost unoccupied cells with $\frac{1}{2}$ . Then fill the two leftmost unoccupied cells with $\frac{1}{2}$ .

Alternate between the left most and right most extremes, till only $1$ cells remains. Fill that cell with any integer.

For example, consider $n = 8$
The sequence would be constructed as follows ( $a$ represents any integer):
$a, , , , , , ,$
$a, , , , , , \frac{1}{2}, \frac{1}{2}$
$a, \frac{1}{2}, \frac{1}{2}, , , , \frac{1}{2}, \frac{1}{2}$
$a, \frac{1}{2}, \frac{1}{2}, , \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}$
$a, \frac{1}{2}, \frac{1}{2}, a, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}$

It is easy to see that this works.

Now, to prove that $2$ is the minimum:
( $x_i$ represents a real number, not an integer and $a_i$ represents an integer)

Obviously, the first/last cell has to be an integer $(= a_0)$ . WLOG, the first cell is an integer. This forces us to fill the last two cells with $x_1$ and $a_1-x_1$ , because the first $2$ cells cannot be an integer unless the second cell itself is an integer. This forces the 2nd and 3rd cell to be filled with $x_2$ and $a_2 - x_2$ , because the sum of the last $3$ cells cannot be an integer unless the third from last cell is an integer.

This process continues, till we are only left with $1$ cell. The remaining $(n-1)$ cells sum to be an integer, as the sum is:
$a_0 + \sum_{j=1}^{i} a_j - x_j + \sum_{j=1}^{i}x_j = \sum_{j=0}^{i}a_j$ which is an integer. Now this last cell has to be filled with an integer, because the sum of all $n$ cells must be an integer.

Hence $2$ is the minimum

This post has been edited 1 time. Last edited by Salty_Titanium, Feb 1, 2018, 7:17 PM

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#4 h Apr 6, 2025, 9:13 PM • 1 Y

Y by PikaPika999

The answer is $2$ . Construction follows $n=4k\Rightarrow a_1=0~a_2=\frac1{1905}~a_3=\frac{-1}{1905}~a_4=\frac1{1905}~a_5=\frac{-1}{1905}... a_{2k-1}=\frac{-1}{1905}~a_{2k}=0~a_{2k+1}=\frac{1}{316972}~a_{2k+2}=\frac{-1}{316972}..... a_{4k}=\frac{-1}{316972}$
$n=4k+2\Rightarrow a_1=0~a_2=\frac{1}{3500}~a_3=\frac{-1}{3500}...a_{2k+1}=\frac{-1}{3500}~a_{2k+2}=0~a_{2k+3}=\frac{1}{90210}~a_{2k+4}=\frac{-1}{90210}...a_{4k+2}=\frac{-1}{90210}$

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