The Absolute's Minimum

Algebra Level 4

$\large |x - 1| + |2x-1| + |3x-1| + \cdots + |300x-1|$

If the minimum value of the expression above is $\dfrac AB$ , and it occurs when $x = \dfrac CD$ , where $(A, B)$ and $(C, D)$ are each coprime pairs of positive integers, find $A+B+C+D$ .

The answer is 13460.

2 solutions

Efren Medallo
Sep 25, 2016

Relevant wiki: Absolute Value Inequalities - 2 Linear Terms

We will make use of the absolute value inequality.

$|x-a| + |x-b| \geq |a-b|$

For every $|kx-1|$ , we split this into $k$ elements of $|x - \frac{1}{k}|$ . Thus, the entire summation would be

$|x-1| + |x- \frac{1}{2} | + |x- \frac{1}{2} | + |x- \frac{1}{3}| + |x- \frac{1}{3} | + |x- \frac{1}{3} | +...$

With $45150$ terms. We now add the corresponding beginning and end terms. That is, the first to the last, the second to the second last, and so on..

From there we get that

$f(x) \geq |1 - \frac{1}{300}| + |\frac{1}{2} - \frac{1}{300}| +...$

Knowing that the last pair constitutes of the 22575th and the 22576th element of the expanded summation, we can find the fraction involved with it.

Notice that $\sum_{n=1}^{211} n = 22366$ , and that $\sum_{n=1}^{212} n = 22578$ . Thus, we know that the middle two elements are $|x - \frac{1}{212}|$ and $|x - \frac{1}{212}|$ . So by the inequality, this should be greater than or equal to zero. For it to be minimum, $x$ has to be $\boxed { \frac{1}{212}}$ .

From there, we can now solve for the minimum value, which is $\frac{13141}{106}$ .

To complete this proof, you need to observe that, for example, the inequality $|x-a| + |x-b| \ge |a-b|$ is in fact an equality for all $x$ between $a$ and $b$ . This will tell us that your final inequality for $f$ is an equality at $x=\tfrac{1}{212}$ .

Mark Hennings - 4 years, 8 months ago

I used the analogy to a graph.Clearly minimum occurs when the graph changes its slope.For some $1/n<x<1/(n+1)$ .One could show that the expression $f(x,n)=2n-300+x(150*301-n(n+1))$ So for the slope to be negative $150*301$ $<n(n+1)$ or $n>=212$ .And for $n=<211$ the slope is positive.So for in the interval $(1/211,1/212)$ the slope is positive and for $x$ in the interval $(1/212,1/213)$ slope is negative.Clearly minimum occurs at $x=1/212$ .Putting the value of $x$ the value of the expression is $13141/106$

Spandan Senapati - 4 years, 1 month ago

great problem and wonderful solution

space sizzlers - 4 years, 8 months ago

Is there any space for calculus here?

neelesh vij - 4 years, 8 months ago

Any other way and how do u know that x-a plus x-b greater than a-b

Tarun B - 4 years, 3 months ago

That's the form of Triangle inequality.Sum of any $2$ sides is greater than third side.Fix three points in complex plane $(x,a,b)$ .Let these form the vertices of a Triangle $ABC$ .Clearly $AB+AC>BC$ . $AB=|x-a|,AC=|x-b|,BC=|a-b|$ .Plugging the values you get the trivial inequality $|x-a|+|x-b|>|a-b|$ .But if the points are collinear then clearly equality holds.

Spandan Senapati - 4 years, 1 month ago

13141/106 is 123. 97168 where as the minimum value of this expression is 123.97058

Aryan Tripathi - 1 year, 5 months ago

Melih Koruyucu
May 11, 2019

|x-1|+|2x-1|+|3x-1|+...+|nx-1|+|(n+1)x-1|+...+|300x-1|

if nth value of the series is 0, then nx-1=0 and n=1/x then it can be writen as

=(1-x)+(1-2x)+(1-3x)+...+(1-nx)+((n+1)x-1)+((n+2)x-1)+...+(300x-1)

=1 n - (x+2x+3x+...+nx) + (300-n) -1 + ((n+1)x+(n+2)x+...+300x)

=2n - 300 - (n (n+1) x/2 + (300-n)(301+n)*x/2

n=1/x as i said before

= 2n - 300 - (n+1)*n/2n + (90300-n-n^2)/2n

= (4n^2 - 600n - n^2 - n + 90300 - n - n^2)/2n

= (n^2 - 301n + 45150)/n

lowest value is where the derivative of the equation becomes zero

d/dn ((n^2 - 301n + 45150)/n) = 1 - (45150/n^2=0

45150/n^2=1

n^2=45150

now when n is 212, equation has minimum value

A/B= minimum value of (n^2 - 301n + 45150)/n = 26282/212 = 13141/106

x=C/D = 1/212

A+B+C+D= 13141+106+1+212 = 13460

The Absolute's Minimum

The answer is 13460.

2 solutions

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