be positive real numbers such that 
Prove that:
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
and write the numbers 
down on a board. A move consists in erasing any two numbers 
and 
,
and 
on the board, and then removing repetitions (e.g., if the board contained the numbers 
,
and 
,
)
with real coefficients of degree 
)![\[
Q(x) \;=\; P(x) \;+\; a\,P'(x) \;+\; a^2\,P''(x) \;+\; \dots \;+\; a^n\,P^{(n)}(x)
\]](http://latex.artofproblemsolving.com/6/5/e/65e5e4eefcadd5d0019b4929b3472cdf203c13aa.png)
has no real roots.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
,
such that 
.
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to 
are collinear, and (iii) 
.
bisects 
.
are chosen uniformly and independently at random on sides 
and 
respectively, of equilateral triangle 
Which of the following intervals contains the probability that the area of 
is less than half the area of 
![$\textbf{(A) } \left[\frac 38, \frac 12\right] \qquad \textbf{(B) } \left(\frac 12, \frac 23\right] \qquad \textbf{(C) } \left(\frac 23, \frac 34\right] \qquad \textbf{(D) } \left(\frac 34, \frac 78\right] \qquad \textbf{(E) } \left(\frac 78, 1\right]$](http://latex.artofproblemsolving.com/5/8/7/5874157b23f9d385f707b696eee3d0780fd8a45b.png)
such that 
is an integer for every positive integer 
be a set of integers with the following properties:
.
and 
,
.
,
is composite, then all positive divisors of 
Y by axsolers_24, kingu, Rounak_iitr
Let 
be integers such that 
. Let 
. For any finite nonempty set 
of positive integers, define ![\[ f(X) = \min_{1 \leqslant i \leqslant n} \sum_{x \in X} \left\{ \frac{x}{a_i} \right\}. \]](//latex.artofproblemsolving.com/9/4/4/9447212baebe95373b95672c72ae5b18d64fcf47.png)
Such a set is called minimal if for every proper subset 
of it, 
always holds.
Suppose is minimal and 
. Prove that ![\[ |X| \leqslant f(X) \cdot M. \]](//latex.artofproblemsolving.com/c/7/3/c73db87bc2d9e3f44853e625625a40aa16510607.png)
be integers such that 
. Let 
. For any finite nonempty set 
of positive integers, define ![\[ f(X) = \min_{1 \leqslant i \leqslant n} \sum_{x \in X} \left\{ \frac{x}{a_i} \right\}. \]](http://latex.artofproblemsolving.com/9/4/4/9447212baebe95373b95672c72ae5b18d64fcf47.png)
Such a set is called minimal if for every proper subset 
of it, 
always holds.Suppose
is minimal and 
. Prove that ![\[ |X| \leqslant f(X) \cdot M. \]](http://latex.artofproblemsolving.com/c/7/3/c73db87bc2d9e3f44853e625625a40aa16510607.png)
This post has been edited 3 times. Last edited by Photaesthesia, Nov 27, 2024, 7:09 AM
compared to the previous two
I wonder if any national team members can solve this during contest.
or 
.
by removing one element each which gives a stronger bound by applying the claim to 
instead of applying to 
.
so suppose 
.
,![\[
\sum_{i=1}^n \frac{x \pmod{a_i}}{a_i} \ge \frac{n + 1}{M}
\]](http://latex.artofproblemsolving.com/5/5/b/55b5bc7c28ec8fd76914e90036493e3af69b6aa1.png)
Proof: Fix 
.
then 
for each 
so this can be verified. Now suppose 
.
which divide 
,![\[
\sum_{i=1}^n \frac{\gcd(a_i, m)}{a_i} \ge \frac{n + d(m) - 1 + 1}{M}
\]](http://latex.artofproblemsolving.com/1/9/f/19f9acbfac02f32a65438b1864f1a2fe409cd4f7.png)
removing all 
so the bound is tightest when there's exactly one term not divisible by ![\[
\frac{\gcd(a, m)}{a} \ge \frac{d(m)}{\text{lcm}(a, m)} + \frac{1}{M} \iff \frac{a(m - d(m))}{a \cdot lcm(a, m)} \ge \frac{1}{M}.
\]](http://latex.artofproblemsolving.com/9/b/a/9ba4482e1ddcf2a8653baf2e4496828acc72626d.png)
(This bound can be tightened a lot I think, which might lead closer to the solution) 
.![\[
\sum_i \sum_{x \in X} \left\{ \frac{x}{a_i} \right\} \ge \frac{n+1}{M} \cdot |X|
\]](http://latex.artofproblemsolving.com/7/7/a/77afed0043d8fa24d8648e0c6ef2f146ba4b8f78.png)
Note that by minimality we get that each 
is within 
of each other. As such, we get that![\[
nf(X) + n \ge \frac{n+1}{M} \cdot |X|
\]](http://latex.artofproblemsolving.com/f/b/5/fb5184b9e85ef1dffcc69633b53a0e1193ff37a1.png)
which becomes the rearranged equality.
,
,
