# 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 $$(T_n)$$?

## One Reply to “INMO 2013 Question No. 4 Solution”

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…..