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INMO 2013 Question No. 4 Solution

 4     Let N be an integer greater than 1 and let \((T_n)\) be the number of non empty subsets S of ({1,2,…..,n}) with the property that the average of the elements of S is an integer.Prove that \((T_n – n)\) is always even.

Sketch of the Proof:

\((T_n )\) = number of nonempty subsets of \(({1, 2, 3, \dots , n})\) whose average is an integer. Call these subsets int-avg subset (just a name)

Note that one element subsets are by default int-avg subsets. They are n in number. Removing those elements from \((T_n)\) we are left with int-avg subsets with two or more element. We want to show that the number of such subsets is even.

Let X be the collection of all int-avg subsets S such that the average of S is contained in S
Y be the set of all int-avg subsets S such that the average of S is not contained in S.

Adding or deleting the average of a set to or from that set does not change the average.
This operation sets up a one-to-one correspondence between X and Y, so X and Y have the same cardinality. Since \((X\cap Y =\emptyset)\), the number of elements in \((X\cup Y)\) is even and hence the number of subsets of two or more elements that have an integer average is even.

Comment

What is the cardinality of \((T_n)\)?

February 8, 2013

1 comment

  1. T_n – n is the number of non-singleton sets ……ar non- singleton sets always occur in pairs,er modhye there exists a S which contains the average.amr mne hoy this is a shorter solution.i was working it out…..

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