An algebra problem by Vishwesh Ramanathan

Algebra Level 2

Find the minimum value of x when it's sum is the least . n varies from 1 to 432.


The answer is 216.

Vishwesh Ramanathan by Vishwesh Ramanathan , 29, India

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3 solutions

Adit Mohan
Feb 17, 2014

first of all the problem is incorrectly stated. what it means is: .
find the value of x for which the expression ∑ n = 1 432 \sum_{n=1}^{432} |x-n| has the minimum possible value.
now simply x=432/2 =216 will give the solution

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what's more: 216 isn't necessary. any real value from 216 to 217(inclusive of both) works

Adit Mohan - 7 years, 3 months ago

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Which value is bigger. 216 or 217. It's clearly stated to find the min value of x.

Vishwesh Ramanathan - 7 years, 3 months ago

exactly

Sayam Chakravarty - 7 years, 2 months ago
Nivedit Jain
Jan 24, 2017

Solved without pen Just thing if X is quite smaller value then sum will be quite large. If X too big again same. So to mimize best option is mid way.

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Sum |x - n| , n= 1,2,...432 ............ = 432x-{ 1, 2 ... 432} = 432x - 432*433/2 =432{x - 216.5)
Minimum is when x = 216. Arithmetic series.

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