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 $latex (X\cap Y =\emptyset)$, 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)$?

3     Let $ (a,b,c,d \in \mathbb{N})$ such that $ (a \ge b \ge c \ge d)$. Show that the equation $ (x^4 - ax^3 - bx^2 - cx -d = 0)$ has no integer solution.

Sketch of the Solution:

Claim 1: There cannot be a negative integer solution. Suppose other wise. If possible $x= -k$ (k positive) be a solution.

Then we have $ (k^4 + ak^3 +ck = bk^2 +d)$. Clearly this is impossible as $ (a\ge b , k^3 \ge k^2 )$ and $ (c \ge d )$.

Claim 2: 0 is not a solution (why?)

Claim 3: There cannot be a positive integer solution. Suppose other wise. If possible x=k (k positive) be a solution.

Then we have $ (k^4 = a k^3 + b k^2 + c k + d)$
This implies that the right hand side is divisible by k which again implies that d is divisible by k (why?).
Let $d=d'k$
Now $(c\ge d) \implies (c \ge d'k) \implies (c \ge k)$.
Thus $(a \ge c \ge k ) \implies (a \cdot k^3 \ge k \cdot k^3 )$.
Hence the equality $ (k^4 = a k^3 + b k^2 + c k + d)$ is impossible.

1.   Let $(\Gamma_1)$ and $(\Gamma_2)$ be two circles touching each other externally at R. Let $(O_1)$ and $(O_2)$ be the centres of $(\Gamma_1)$ and $(\Gamma_2)$, respectively. Let $(\ell_1)$ be a line which is tangent to $(\Gamma_2)$ at P and passing through $(O_1)$, and let $(\ell_2)$ be the line tangent to $(\Gamma_1)$ at Q and passing through $(O_2)$. Let $(K=\ell_1\cap \ell_2)$. If KP=KQ then prove that the triangle PQR is equilateral.

Discussion:

We note that $(O_1 R O_2 )$ is a straight line (why?)
Also $ (\Delta O_1 Q O_2 , \Delta O_1 P O_2 )$ are right angled triangles with right angles at point Q and P respectively.
Hence $ (\Delta O_1 Q K , \Delta O_2 P K )$ are similar (vertically opposite angles and right angles)
Thus $ ( \frac{KP}{KQ} = \frac{O_2 P} {O_1 Q} = 1)$ as KP =KQ.
Hence the radii of the two circles are equal..
This implies R is the midpoint of $ (O_1 O_2)$ hence the midpoint of hypotenuse of  $ (\Delta O_1 Q O_2)$
$ (O_1 R = RQ = R O_2)$ since all are circum-radii of $ (\Delta O_1 Q O_2)$.
Hence  $ (\Delta O_1 Q R)$ is equilateral, similarly $ (\Delta O_2 P R)$ is also equilateral.
Thus $ (\angle PRQ)$ is $ (60^o)$ also \( RQ = O_1 R = O_2 R = RP \).
Hence triangle PQR is equilateral.