Find the smallest positive integer $n$ for which $|x − 1| + |x − 2| + |x − 3| + · · · + |x − n| \geq 2022$ for all real numbers $x$.
I don't think I can combine any of these terms, right? So I started by changing the equation into the sum of an arithmetic series, but I don't think that does anything.
$$\sum_{i=1}^n |x-i| = \frac{n}{2}(|x-1|+|x-n|) \ge 2022$$
I'm not sure how to go from here, any suggestions? Thanks!
Hint: to minimize the sum, you want an $x$ that minimizes the distance from $1, 2, \dots, n$: intuitively, what does this $x$ have to be?
It should be \begin{align} x &= \frac{n+1}{2}, \end{align} the average of $1, 2, \dots, n$.
Then, compute the value of the sum at this $x$.
For $n = 2k + 1$, the sum is $2 \sum_{m=1}^k m = k(k+1)$. For $n = 2k$, the sum is $2 \sum_{m=1}^{k} \left(m - \frac{1}{2}\right) = k^2$.
Then compute when those values are $\geq 2022$.
In particular, since $44 \cdot 45 < 2022 < 45^2$, we see that we should pick $n = 90$.
The expression given takes the smallest of its values when x is chosen to be $\frac{n}{2}$. We care about its smallest value since we want $n$ to have the smallest value which makes the expression greater than or equal to $2022$.
For odd $n$, we have $x = \frac{n+1}{2}$ and for even $n$, we have $x = \frac{n}{2}$. This will give us a sum of $$\left(\frac{n}{2}-1\right)+\left(\frac{n}{2}-2\right)+\left(\frac{n}{2}-3\right)+...+2+1+0+1+2+3+...+\left(\frac{n}{2}-1\right)+\left(\frac{n}{2}\right)$$ for any even $n$.
Since we have $n$ terms the summation becomes $$\frac{n^2}{2}-2\cdot\left(\frac{n}{2}\cdot\frac{\frac{n}{2}+1}{2}\right)+\frac{n}{2}=\frac{n^2}{2}-\left(\frac{n^2+2n}{4}\right)+\frac{n}{2}=\frac{n^2}{4} \geq 2022$$ $+\frac{n}{2}$ is because we subtracted $\frac{n}{2}$ twice by multiplying $\left(\frac{n}{2}\cdot\frac{\frac{n}{2}+1}{2}\right)$ by $2$, but it occurs only once.
