I stumbled upon an interesting problem and I wonder whether you see any interesting alternative solutions. I am particularly interested in the method that allows to determine the number of possible cases.
The problem is the following:
Find all values of $l$ that satisfy the equation below:
$|x-1|+|x-2|+|x-3|+⋯+|x-2019|+|x-2020|≤l$
So currently I am using the following absolute value property:
$|a|+|b|+⋯+|n|≥|±a±b±⋯±n|$
With this I consider the two cases, where the x simplifies (I show only one case for simplicity - they have the same resulting value):
$|x-1|+⋯+|x-2020|≥|(x-1)+⋯+(x-1010)-(x-1011)-…-(x-2020)|$
$|x-1|+⋯+|x-2020|≥|-1-2-3-…-1010+1011+⋯+2019+2020|$
$|x-1|+⋯+|x-2020|≥1020100$
$l∈[1020100,∞[$
And now I am wondering. Do you see any other method to solve the equation, apart of the property I showed above?
I know there are these methods with considering different cases. But how many cases are there? Is it 2021?
I am waiting for your opinions :)
The function is piecewise linear, with a slope regularly increasing from $-2020$ to $2020$, in steps $2$. Thus it has a flat single minimum which is reached on the symmetry axis, in $[1010,1011]$. The function value is indeed
$$1009+1008+\cdots 1+0+1+2+\cdots1010=1020100.$$
To illustrate, the function with $6$ instead of $2020$.