AoPS Academy
cards on which the numbers 
, 
, 
, are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let 
be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer 
for which Bob can guarantee that is at least .
such that
for any positive integers 
, 
.
, with a,b,c,d are arbitrary integers and p is an arbitrary prime such that the fraction is an integer too.
and 
be positive integers, and let 
be a 
grid of unit squares.
or 
rectangle. A subset of grid squares in is domino-tileable if dominoes can be placed to cover every square of exactly once with no domino extending outside of . Note: The empty set is domino tileable. to the upper-right corner of formed by exactly 
edges of the grid squares. and , the number of up-right paths that divide into two domino-tileable subsets.
such that
whenever 
is defined. What is 
+1 w
and 
be positive integers. Prove that there exists a positive integer such that for every odd integer 
, the digits in the base-
representation of 
are all greater than .
and be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
which is obtained with 
. We now show this is maximal.![\[
x_1 \le x_2 \le x_3 \dots \le x_{2024}
\]](http://latex.artofproblemsolving.com/9/a/e/9aee904e525006c930cb8a966552f506dbfb8650.png)
We first claim that we may smooth such that 
. First, if any element is more than 
, we may smooth that number to be at no cost. Now, the idea is that if we have minimal elements equal to 
and 
maximal elements equal to 
, we may increase the minimal elements and decrease the maximal elements until the number of minimal / maximal elements changed. Formally, our constraint is that![\[
\frac{M\varepsilon}{m} + (x_{2024} - e) - x_{2024}^2 \ge 0
\]](http://latex.artofproblemsolving.com/f/b/f/fbf2dd82a119315578643ee97bc24ffe14ba53cc.png)
which holds for 
. Firstly, if 
then the maximal number of pairs is at most 
so we are done, and thus assume 
. Next, if 
and 
, then the maximal number of pairs is at most 
and thus we are done. As such, we may assume in what follows that 
. Since 
, this means we can repeatedly smooth until all elements more than 
are equal to it. elements equal to , elements at least 
less than , and 
remaining elements. Then we get at most 
pairs since for a pair 
that works, either 
is an or element, or 
is an element and is a element. Then, we may note that 
holds except for the cases 
or 
. In the latter case, one element is and the remaining 
are 
, so there are 
pairs.
. Then there are 
elements which are more than 
, and we can upper bound with 
so we are done.
Prove that
non-negative real numbers 
that satisfy 
,![\[
x_i^2 + x_j \geqslant \frac{1}{2023}.
\]](http://latex.artofproblemsolving.com/2/3/c/23c312912bb0804a499fa33d1e254c3c1b554fad.png)
