An average problem

Probability Level 5

The natural numbers from 1 to 2015 are placed on a line in any order, so that no two numbers lie on the same point and that their average mean does not lie between those two numbers.

For example: You place the 1 at the very left, 2015 at the very right. This leads to a problem because $(1+2015) \over 2$ = 1008 is their average and has to lie between 1 and 2015., therefore those two numbers would both not suffice the property.

However one could place numbers in the following order 2015 1 2 3 ..... 2013 2014. Now 2015 would suffice the property stated above. However the numbers 1 and 3 now make a problem since the number 2 lies between them..

Is it possible to find an arrangement of those numbers, so that all numbers suffice the property given above? If not, what is the maximum amount of numbers to hold this property?

The answer is 2015.

1 solution

Kenny Lau
Jul 22, 2015

Allow me to rewrite the question to make it clearer.

Pick natural numbers from $1$ to $2015$ without repeating, such that the mean of the numbers picked is not between any two numbers picked.

Find the maximum numbers picked.

It is obvious that the solution is to just pick all the numbers, then the mean would be $\dfrac{1+2015}2=1008$ which is also a number picked.

There are $\fbox{2015}$ numbers from $1$ to $2015$ .

But, the question remains of what arrangement of the numbers $1,2,\cdots,2014,2015$ would satisfy the 'no mean in between' criterion.

Janardhanan Sivaramakrishnan - 5 years, 10 months ago

I see. I came too early. The question was highly modified after I posted this solution, that my solution is no longer valid due to my misunderstanding of this question.

Kenny Lau - 5 years, 10 months ago

Can you explain in detail?

Hari prasad Varadarajan - 5 years, 10 months ago

An average problem

The answer is 2015.

1 solution

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