AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, we know that lines 
and 
intersect at 
, lines 
and 
intersect at 
, and lines 
and 
intersect at 
. Suppose that the circumcircle of 
intersects line 
at 
and 
, and the circumcircle of 
intersects line at 
and 
, where 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
, then 
.
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
, and let be the midpoint of the side . Let be a point on the side and a point on the side such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle .
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
contestant of EGMO are named 
. After the competition, they queue in front of the restaurant according to the following rules.
with 
.
has at least other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly positions. has fewer than other contestants in front of her, the restaurant opens and process ends. the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves.
numbers 
and 
numbers 
are written in some order.
.
, where 
is a prime number. Show that if is an even number, then the polynomial ![\[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]](http://latex.artofproblemsolving.com/5/d/f/5df80f885e63852e4f53a6145dafd18d63dc3526.png)
is divisible by 
.
reals, and 
.
?
be a triangle with 
. Let be the orthocenter of . Point is the midpoint of and point lies on the side such that 
. Prove that
.
be a regular polygon, and let 
be its set of vertices. Each point in is colored red, white, or blue. A subset of is patriotic if it contains an equal number of points of each color, and a side of is dazzling if its endpoints are of different colors. is patriotic and the number of dazzling edges of is even. Prove that there exists a line, not passing through any point in , dividing into two nonempty patriotic subsets. be an integer greater than 
. For a positive integer 
, let 
. Suppose that there exists a 
-element set 
such that is an -element subset of 
; shares at most one common element; is contained in exactly two elements of . in terms of .
eight-digit positive integers that use each of the digits 1, 2, 3, 4, 5, 6, 7, 8 exactly once. Let N be the number of these integers that are divisible by 
. Find the difference between 
and 2025.
and get 
, which gives us a final answer of 
However, 
same issue wtih 
Thus both alternating set of digits sums to 
ways.
, so it is impossible for 3 digits to be in either one of ![$[1, 4], [5, 8]$](http://latex.artofproblemsolving.com/5/d/9/5d9f6f78361cd851c801b6bcd5a328c8ec93bc07.png)
by symmetry. Thus it is split 2-2. We do casework on the mean of the smaller 2, since the mean of the larger 2 must then symmetricly be 
.
we have 1 case. 
gives 1 case. 
gives 
cases for each pair so 
total. 
both give 1 case. Note that in each case, 2 numbers are even and 2 are odd. So for each of the 
cases there are 
ways to order.
, we simply want 
and 
is a multiple of 2
, if 
, but 
is not achievable since 
and you reverse the elements in each pair. be a multiples of 2, there are in total 
ways, the answer is then 
such that 
, 
, 
, 
, 
, and the sums of 
and differ by a multiple of 
.
such pairs. We claim that 
. even digits we can put at the end. Then, by the divisibility rule for , there are 
ways to make and take up alternating digits.
, so our claim is proven.
.
, we want the sum of to be 
or 
. Since 
, the sum of must be .
, so our answer is .
, so the individual sums of each alternating sequence is either 
or 
. Only the latter is possible, now we carefully list partitions that sum to with the given digits:![\[ 9621 \quad 9531 \quad 9432 \quad 8721 \quad 8631 \quad 7631 \quad 7542 \quad 6543 \]](http://latex.artofproblemsolving.com/4/d/6/4d650ceac5f8e5c6f7d0da0257be50d3909536e3.png)
There are partitions. Now the units digit is even so the alternating sum with the units digit has 
possibilities, and we can arbitrarily permute the other alternating sum ways. The requested answer is![\[ \left| 8 \cdot \dfrac{4!}{2} \cdot 4! - 2025 \right| = \left|2304 - 2025 \right| = \boxed{279}. \]](http://latex.artofproblemsolving.com/4/4/1/44137544dcfd094ca4e0a929b3712b5fd1eeeaff.png)
counting a total of 4 pairs. Notice now, the arrangement of the odd-positioned numbers doesn't matter so we can say it is 4! or 24 ways. The arrangement of each of these new 4 pairs we have just found is 3! = 6. Thus, for the unit's digit to be 2, we have 24*6*4 possible ways. Since we claimed that this was symmetric over the rest of the even digit unit's digits, we have to multiply this by 4 again. So our total number of ways is 24*6*4*4 = 2304. Thus, the positive difference between N and 2025 is 2304 - 2025 = .
be the number, and 
, and 
.
, if and only if 
. Since 
and 
, we easily see that only 
works., are![\[
(1, 2, 7, 8), (1, 3, 6, 8), (1, 4, 5, 8), (1, 4, 6, 7), (2, 3, 5, 8), (2, 3, 6, 7), (2, 4, 5, 7), (3, 4, 5, 6)
\]](http://latex.artofproblemsolving.com/d/8/0/d80edc38c068442f6fb9700f7adada8dc3c3f222.png)
and their permutations. Each even, and odd numbers. Since 
must be even, for each one above, there are 
combinations for 
, and 
for 
. The number of such are 
. In particular, the answer is 
.
and 
. Prove that
. Prove that
Thank lbh_qys.
be the sum of even positions and 
be sum of odd positions. Note 
but 
and we may see that only 
works. We find quadruplets of four digits which sum to 
and similarly four odds fails so by parity we must have two evens and two odds to sum to ![\[8 \cdot 12 \cdot 24 = 2304 = 2025 + \boxed{279}\]](http://latex.artofproblemsolving.com/d/9/9/d99176ce8e4fbbe8983f70e03162c0f100fe2c40.png)
extraction
and 
.
