MHB What is the least value of the sum $|p-1|+|p-2|+\cdots+|p-10|$?

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The least value of the sum |p-1| + |p-2| + ... + |p-10| occurs when p is the median of the numbers 1 through 10, which is 5.5. This minimizes the total distance from p to each integer in the set. The minimum value of the sum is calculated to be 25. The problem highlights the concept of minimizing absolute deviations in a set of numbers. Understanding this principle is crucial for solving similar optimization problems.
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Determine the least value of the sum $|p-1|+|p-2|+\cdots+|p-10|$ where $p \in R$.
 
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Nice question, here is a solution
I think p=5.5

|------|------|------|------|--\/--|------|------|------|------|
1...2...3...4...5.,5.5,..6...7...8...9...10

The sum equal in this case 9+7+5+3+1 = 25
 
anemone said:
Determine the least value of the sum $|p-1|+|p-2|+\cdots+|p-10|$ where $p \in R$.

Hello.

by symmetry:

If it were 9 elements:

4+3+2+1+0+|-1|+|-2|+|-3|+|-4|=20

Is now equivalent to insert "5" or "- 5"

Solution: p \in{ } [6,5]

Regards.
 
$Let \; \; S(p) = \sum_{i=1}^{10}\left | p-i \right |= \sum_{i=1}^{10}\sqrt{(p-i)^2} \\\\ S'(p)= \sum_{i=1}^{10}\frac{(p-i)}{|p-i|}= (\pm) 1(\pm) 1...(\pm) 1$,
where each sign depends on the sign of the difference: $p-i$.
The minimal sum requires: $S'(p)=0$, which happens when the signed ones annihilate:

$\sum_{i=1}^{10} (\pm )1 = 0.$ (Therefore, the case $p=i$ is not of interest here)

Thus, there must be exactly five $+1$ and five $-1$. Therefore $p$ is the midpoint between $5$ and $6$

And the minimal sum is: $S_{min}=2*(5*5.5-15)=25$.
 
Thank you so much for participating, guys! :)

Solution taken from other intelligent mind:
[FONT=MathJax_Main]|
Note that $|p-m|+|p-(11-m)| \ge 11-2m$. If we are to add the inequalities for $m=1,\,2,\,3,\,4,\,5$, we get $|p-1|+|p-2|+\cdots+|p-9|+|p-10| \ge 55-2(1+2+3+4+5)=25$, hence, the least value of the specified sum is 25.
 
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