practice problems pending

Q1.

(n) is a positive integer.

(A) is a set such that A={1,2,...,n}.

Let (t_n) denote the number of subsets of (A) such that the arithmetic mean (AM) of the elements is an integer.

Prove that (t_n) and (n) are both odd or both even.

S1.
1. Let us ignore empty subset since A.M. is not defined for that.
2. Subsets of size 1: {1}, {2} ... {n}. There are 'n' such subsets. And each of them has their A.M. as integer.
3. Now let's consider the subsets with size >= 2 which have an integer A.M.
Let G be the set of all such subsets.
Let's consider a function 'f' defined for subsets of size >= 2.

f(S) = S - {k} where k is the A.M. of S and k is present in S.
f(S) = S + {k} where k is the A.M. of S and k is not present in S.

For e.g.
S = {1,3,8}, k = (1+3+8)/3 = 4
4 is not there in S.
f(S) = {1,3,4,8}

S = {1,2,3}, k = (1+2+3)/2 = 3
3 is there in S.
f(S) = {1,2}

Note 1: S and f(S) have the same A.M.
Why?
Let S = {a1,a2...,ak}
AM = sigma(a_i)/k
If you remove AM, the new AM = 
[sigma(a_i) - sigma(a_i)/k]/(k-1) = (k-1)*sigma(a_i)/k*(k-1) = sigma(a_i)/k = same as before

If you add the AM, the new AM = 
[sigma(a_i)  + sigma(a_i)/k]/(k+1) = (k+1)*sigma(a_i)/k*(k+1) = sigma(a_i)/k = same as before

Note 2: Removing an element from S doesn't reduce its size to 1.
Why?
It will happen only if the size of S is 2.
Let S = {a,b}, AM = (a+b)/2
For 2 distinct numbers AM can't be same as any number.
So, for AM to be removed, S needs to have at least 3 numbers.


Note 1 and 2 taken together prove that S and f(S) both belong to G.

Note 3:
f(f(S)) = S
Why?
Case 1:
If f(S) removed the AM from S, as we saw, it still has the same AM.
So f(f(S)) will again add the AM since it's missing.
So we get 'S' back again.

Case 2:
If f(S) added the AM, f(f(S)) will remove the AM to give us back 'S'.

Note 4:
f(S) != S
Why?
Since they have different sizes, they can't be same.

Note 3 and 4 taken together prove that the mapping f(S) divides G into 2 mutually exclusive subsets which have one to one mapping.

So number of elements in G is even.

t_n = n + an even number
=> t_n and n have same parity.



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