Find the number of positive integer pairs $(x, y, z)$ that $$\frac{1}{x+1}+\frac{1}{y+2}+\frac{1}{z+3}=\frac{11}{12}$$

$99$ different points $P_1, P_2, ..., P_{99}$ are marked on circle $O$. For each $P_i$, define $n_i$ as the number of marked points you encounter starting from $P_i$ to its antipode, moving clockwise. Prove the following inequality. $$n_1+n_2+\cdots+n_{99} \leq \frac{99\cdot 98}{2}+49=4900$$

Acute triangle $ABC$ satisfies $\angle A > \angle C$. Let $D, E, F$ be the points that the triangle's incircle intersects with $BC, CA, AB$, respectively, and $P$ some point on $AF$ different from $F$. The angle bisector of $\angle ABC$ meets $PQR$'s circumcircle $O$ at $L, R$. $L$ is the point closer to $B$ than $R$. $O$ meets $DF, DR$ at point $Q(\neq F, L), S(\neq R)$ respectively, and $PS$ hits segment $BC$ at $T$. Show that $T, Q, L$ are collinear.

find all positive integer n such that there exists positive integers (a,b) such that (a^n + b^n)/n! is a positive integer smaller than 101

$ABC$ is a right triangle with $\angle C$ the right angle. $X$ is some point inside $ABC$ satisfying $CA=AX$. Let $D$ be the feet of altitude from $C$ to $AB$, and $Y(\neq X)$ the point of intersection of $DX$ and the circumcircle of $ABX$. Prove that $AX=AY$.

Find all pairs $(n, p)$ that satisfy the following condition, where $n$ is a positive integer and $p$ is a prime number. Condition) $2n-1$ is a divisor of $p-1$ and $p$ is a divisor of $4n^2+7$.

Let $A_k$ be the number of pairs $(a_1, a_2, ..., a_{2k})$ for $k\leq 50$, where $a_1, a_2, ..., a_{2k}$ are all different positive integers that satisfy the following. $\cdot$ $a_1, a_2, ..., a_{2k} \leq 100$ $\cdot$ For an odd number less or equal than $2k-1$, we have $a_i > a_{i+1}$ $\cdot$ For an even number less or equal than $2k-2$, we have $a_i < a_{i+1}$ Prove that $A_1 \leq A_2 \leq \cdots \leq A_{49}$.

$f$ is a function from the set of positive integers to the set of all integers that satisfies the following. $\cdot$ $f(1)=1, f(2)=-1$ $\cdot$ $f(n)+f(n+1)+f(n+2)=f(\left\lfloor\frac{n+2}{3}\right\rfloor)$ Find the number of positive integers $k$ not exceeding $1000$ such that $f(3)+f(6)+\cdots+f(3k-3)+f(3k)=5$.