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How many solutions has an equation $$|x-1|+|x-2|+|x-3|+...+|x-2002|=a$$depending on an a parameter? In my opinion, an equation can have 0, 2 or infinite number solutions, but I don't know how to prove it.

  • In my opinion, an equation can have 0, 2 or infinite number solutions, but I don't know how to prove it. –  May 30 '19 at 11:16
  • First, what is common term? Can you write complete equation for some small number of terms? Second, try to draw plot of left part, it will probably give you a hint. – mihaild May 30 '19 at 11:18
  • I know it, but it is hard to prove that the plot will have such shape. –  May 30 '19 at 11:19
  • What is common term? It's really confusing to see first subtrahend been first natural numbers, but the last been $2$ again. – mihaild May 30 '19 at 11:22
  • Sorry, fixed the question. –  May 30 '19 at 11:24
  • What have you tried? What do you think the answer is? – Parcly Taxel May 30 '19 at 11:26
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    Most of what you've written in the comments should be edited into the question text instead. What you have tried and what you think are essential parts of the question, don't rely on people to read the comments to find out. – David K May 30 '19 at 11:29
  • The left hand side is a convex function. You can use a tool like the Three Slope Lemma (or a direct proof) that if it takes the value $a$ at $3$ points $x < y < z$, then it will take the value $a$ on the interval $[x, z]$, i.e. infinitely many points. Your opinion is correct. – Theo Bendit May 30 '19 at 11:31
  • If $x$ is a solution then clearly $2003-x$ is also a solution, so you can exclude the possibility that there are an odd number of solutions, except in the case of $x=\frac{2003}{2}$ being a solution and in that particularly case all $x \in [1001,1002]$ give the same sum – Henry May 30 '19 at 11:37
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2 Answers2

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You may take the RHS (call it $f(x)$) as 2002 times mean deviation of first 2002 natural numbers about $x$. The mean deviation is known to be least if measured about the median. The medians of this data being 1001 and 1002. So $f(x)$ admits minimum value at every point on $x \in[1001,1002].$ So if $a=f[1001]=f[1002]=1002001$ this equation has infinitely many roots in $[1001,1002].$ If $a< 1002001$. The equation does not have a root. For $a>1002001$ the equation has exactly two real roots. Graphically $f(x)$ is an open polygon going to infinity on both sides of its minimum and it is constant (the minimum) in the domain [1001,1002].

Z Ahmed
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The function on the left measures the total distance from the point $x$ to $2002$ other points. If $x\lt 1$ it is to the left of all the points and shifting to the right by $\epsilon$ reduces the sum by $2002\epsilon$.

In fact if there are $n$ points to the left of $x$ and $2002-n$ points to the right, increasing $x$ by $\epsilon$ increases the function by $n\epsilon-(2002-n)\epsilon=2\epsilon (n-1001)$. So while $n\lt 1001$ the function is decreasing, while $n=1001$ it is static and for $n\gt 1001$ it is increasing.

[Analysis of what happens at the actual points is trivial - $n$ changes and the behaviour is clear. It is obvious that the function is continuous].

So here the minimum possible value of $a$ for which there is a solution is given at the points for which $n=1001$ (and the endpoints of that interval). The function increases without limit, so any greater value of $a$ is achieved at two points.

Note that this analysis does not depend on the fact that the points have integer spacing - it just depends on the number of points either side of $x$.

Some people might want to fill in technical details, but I hope this helps.

Mark Bennet
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