Too much of absolute

Algebra Level 3

What is the minimum value of

x 1 + 2 x 1 + 3 x 1 + . . . + 119 x 1 |x-1|+|2x-1|+|3x-1|+...+|119x-1|

Where x x is a real number.

Note : . |.| denotes absolute value function

Request : please contribute to this


The answer is 49.

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1 solution

Because of the expressions within the absolutes, the minimum will be at a fraction between 0 and 1 of the form 1 n \frac1n where n is a positive integer 119 \leq 119 . By direct computation that minimum is 49 at both 1 85 \frac1{85} and 1 84 \frac1{84} . Further examination shows that the expression is 49 between those fractions.

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