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Could you please assert the last line more clearly?

The numbers of irreducible fractions is even. For example: if $n=6$,

$\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6}$. Here, $\frac{1}{6}$ and $\frac{5}{6}$ is irreducible and there number is tow, in fact even.

aritra barua wrote:Let us denote by $ABC$,the area of $\bigtriangleup ABC$

Use $[ABC]$ to denote the area of $\triangle ABC$. It is more natural and used frequently.

The proof is trivial for some $n$ which has the form $2k$ where $k$ is an integer.Now,to say,this problem is equivalent to proving that the number of reducible fractions is even when $n$ is odd because the aforementioned sequence has precisely 'n-1' fractions(I.g:even numbered fractions).Now,we take $n$ as composite because there is nothing to prove when $n$ is a prime.Let us take a reducible fraction $\frac{n-i}{n}$ when g is a common divisor of 'n' and 'n-i'.This means that g|I and thus there is another fraction which is reducible $\frac{i}{n}$ exclusive of $\frac{g}{n}$ and $\frac{n-g}{n}$.But that follows that every time we choose a reducible fraction,there would be another reducible fraction and thus the number of reducible fractions will be even.It also says that the number of irreducible fractions will also be even.

$\text{Problem 16}$

There are $n$ ants on a stick of length one unit, each facing left or right. At time $t = 0$, each ant starts moving with a speed of $1$ unit per second in the direction it is facing. If an ant reaches the end of the stick, it falls off and doesn’t reappear. When two ants moving in opposite directions collide, they both turn around and continue moving with the same speed (but in the opposite direction). Show that all ants will fall off the stick in at most $1$ second.

As the ants are arbitrary, when they collide and move in opposite directions, basically nothing changes. For example, let's name them A and B. A was going right and B left. When they collide, they change directions, so A goes left and B goes right. But as the ants are similar, we can just take it, that they are continuing in the orignal path.
So, now, the longest path an ant can cross is if it is on one end of a stick and going in the opposite direction. Then, the stick is 1 unit long and the speed of the ant is 1 unit per second.
So, the max time the ants will take to fall off the stick is 1 second.

Problem $17$ : Let $\bigtriangleup$ $ABC$ be an acute triangle.$D$ is the foot of perpendicular drawn from $C$ on $AB$.Let the bisector of $\angle$ $ABC$ intersect $CD$ at $E$ and $\bigcirc$ $ADE$ at $F$.If $\angle$ $ADF$ =45˚,prove that $CF$ is tangent to $\bigcirc$ $ADE$.

aritra barua wrote:Problem $17$ : Let $\bigtriangleup$ $ABC$ be an acute triangle.$D$ is the foot of perpendicular drawn from $C$ on $AB$.Let the bisector of $\angle$ $ABC$ intersect $CD$ at $E$ and $\bigcirc$ $ADE$ at $F$.If $\angle$ $ADF$ =45˚,prove that $CF$ is tangent to $\bigcirc$ $ADE$.

My solution:

I think you should give the full solution to the problems or give hints if the thread is stuck for some days.

OKAY PEOPLE!! Time to make this thread hot again.
Problem $18$
You and your spouse invited $10$ couples to a party. In that party, some of them shook hands with others. But no one shook hands with their spouse. After the party, you asked everyone(including your wife) how many person they shook hands with. You found that everyone shook hands with a different number of people. How many people did your wide shake hand with?