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They start, from Mirpur, one with the bike($ A$) and the other($ B$) on foot, and from Lalmatia($ C$) on foot of course. After one hour $ A$ will give the bike to $ C$, and $ B$ will stop, and wait for the bike. $ A$ will reach Lamatia after 2 hours. $ C$ rides the bike until he meets $ B$ , and gives it to him. After this they'll reach their targets in one hour, so $ 2$:$ 40$ in total.

Problem $23$

Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.

$\text{Problem 23}$

$\text{Problem 24}$
Let $n$ be a positive integer and let $a_1, a_2,.....a_k$(here $k$ > 1) be distinct integers in the set {${1,2.....n}$} such that $n$ divides $a_i(a_{i+1}-1)$ for $i = 1,2,.....k-1$. Prove that $n$ does not divide $a_k(a_1 - 1)$

Solution to problem $24$ (OVERKILL)

We prove by contradiction.
Assume that $n$ divides $a_k(a_1-1)$.

Assume that $(a_i, n)=1$. Then $n| a_i(a_{i+1}-1)$ implies $n|a_{i+1}-1$ which implies $a_{i+1}=1$. Then $n|1(a_{i+2}-1)$ which is only possible if $a_{i+2}=1$ but this contradicts the distinction.
So there exist some prime $p$ such that $p|n$ and $p|a_i$.
Now we inductively prove that $p$ divides all $a_j$ such that $1\leq j \leq n$. We have already shown the base case. Now assume that $p|a_m$. We have, $p |n |a_{m-1}(a_m-1)$. So, we must have $p|a_{m-1}$. Here, the indices are taken modulo $k$. So, $p$ divides all $a_j$ such that $1\leq j \leq n$. Note that the induction was done by starting from $a_i$. We get that if some prime divides at least one of the $a_j$'s, that prime divides all the $a_j$'s.
We will strengthen the statement more with dealing with prime powers.
Assume that $p^x||n$. Then,as $p^x||n|a_j(a_{j+1}-1)$ for all $j$, $p^x|a_j$ for all $j$.
So, we get that $(n,a_1)=(n,a_2)=.......=(n,a_k)= C$ where C is an integer greater than $1$.
Let, $b_j=\frac {a_j}{C}$ for all $j$. Also let, $n_1=\frac {n}{C}$. Then, we have $(n_1, C)=(n_1,b_j)=1$ for all $j$.
Now, as $n_1C|b_jC(b_{j+1}C-1)$, we get $n_1|b_{j+1}C-1$ for all $j$. So, $n_1|b_xC-1-b_yC+1=C(b_x-b_y)$.
We get that $n_1|b_x-b_y$. Which is impossible since $b_x, b_y$ are different integers less than $n_1$. So we get a contradiction. So our first assumption is false. Now, this solution is really unclear. Someone tell me where does the logic breaks if our first assumption is false.

Solution to Problem $24$ (NOT OVERKILL)

As the previous solution we solve by contradiction.
So, assume that $a_1a_k \equiv a_k (mod n)$.
$ a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1\cdot\dots\cdot a_{k-2}\cdot a_k\equiv\dots\equiv a_1\cdot a_{k}\equiv a_k$.

Therefore, $ 0 < |a_1-a_k| < n$ is divisible by $ n$. Contradiction.
EID MUBARAK

As this is the Beginner's Marathon, I request everyone to not give shortlist problems.

Problem $25$

Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.

Let $AD \cap \Omega = P$. Now, note that $AP = BB_1, AA_1 = CA_2$ and $BB_1 = CB_2$. Now, power of point implies that $AP.AD = AC.AA_1 \Rightarrow BB_1.BC = CA_2.AC \Rightarrow CA_2.AC=CB_2.BC$. Therefore, $A,B,A_2,B_2$ are cyclic.

Somebody post the next problem.

Problem $26$
Given an acute angled triangle $\bigtriangleup ABC$.Inscribe in it a triangle $UVW$ of least possible perimeter.