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.
people be arranged in a circle, numbered clockwise from 
to 
. These people are eliminated one by one in order, until only one person remains. The elimination follows this rule: among the remaining people, start counting clockwise from the person with the smallest number, and eliminate the -th person in that count. Then, among the remaining people, start counting again from the person with the smallest number and eliminate the -th person. Repeat this process until only one person remains. Let 
denote the number of the last remaining person.
, people are eliminated in the following order: 
. Thus, 
. It is known that 
under certain conditions. Prove that the necessary and sufficient condition for this is that both 
and 
are prime numbers.
be the unit circle in the complex plane. Let 
be the map given by 
. We define 
and 
for 
. The smallest positive integer such that 
is called the period of 
. Determine the total number of points in 
of period 
.
)
and 
and 
consecutive even numbers, such that the sum of their squares divides the square of their product.
. Solve the equation: 
be a triangle and let 
be its circumcenter and 
its incenter.
be the radical center of its three mixtilinears and let 
be the isogonal conjugate of .
be the Gergonne point of the triangle .
is parallel with line 
.
be a positive integer. Prove that there exist three permutations 
, 
, and 
of 
such that ![\[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]](http://latex.artofproblemsolving.com/9/8/0/98097fb721722b9103ef65d72322956877e262cb.png)
for every 
.
, no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers 
(
is allowed) such that 
. Find the minimal .
and a point , let 
be the cevian triangle of with respect to . Let 
be the orthocenter of , and denote the isotomic conjugate of 
with respect to by 
, respectively. Let the centroid of be 
, and denote the isogonal conjugate of with respect to by 
. Prove that
or in brief
be the circumcenter and the orthocenter of an acute triangle . Prove that the area of one of the triangles 
, 
and 
is equal to the sum of the areas of the other two.
line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands 
times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time. is odd. is even.
such that a positive integer power of 
equals to a positive integer power of 
.
can be written in the form
where 
for some non-negative 
such that
for every 
.
to the following system:
, 
, and![\[
a_{2k+1} = 2 + 2a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/c/4ec892df19b4ec97af9bb1e5d1954e6c26be3172.png)
for all integers 
. Determine all positive integers 
such that![\[
\frac{a_n}{n}
\]](http://latex.artofproblemsolving.com/3/d/1/3d1d012e7c41e1e09abd14626a79b1fb1cbab34d.png)
is an integer.
is odd. Now calculate some values for 
you can see that for 
type numbers 
is an integer. Then just simple induction proves the result.
, , and![\[
a_{2k+1} = 2 + a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/d/4edb78766605f6eb84f2822706527ef16b0e11ef.png)
for all integers . Determine all positive integers such thatis an integer.
, , andfor all integers . Determine all positive integers such thatis an integer.
for all , then only possible integer solution for 
is or 
. is obvious , due to the solution present by others, also values of new sequence you give will be smaller than or equal to the original sequence.
Hatt's off to the proposer of Problem 6 in INMO 2025. The only problem I fail to solve in the whole paper. I was near but didn't observe certain things. I only solved the case when 
but couldn't go further and finally saw the solution in aops.
for all 
. The following is the key claim.,![\[n<a_n \le 2n\]](http://latex.artofproblemsolving.com/d/0/b/d0b8b50120725a3f797b06c6cd527bdb785a366a.png)
trivially, 
and 
so the base cases are clear. Now, say the claim holds for all 
for some 
. Then,![\[2k+1 < 2+2k<2+2a_k=a_{2k+1} = 2+2a_k \le 2+2(2k)=2(2k+1)\]](http://latex.artofproblemsolving.com/1/6/6/16634888e5890ded272d675dba07d59bc8178b0a.png)
with equality on the upper bound if and only if 
. Similarly,![\[2k+2 < 2+k + (k+1) < 2 + a_k + a_{k+1} = a_{2k+2} = 2 + a_k + a_{k+1} \le 2+2k + 2(k+1)= 2(2k+2)\]](http://latex.artofproblemsolving.com/b/1/b/b1bd3d3247e313fbcd9ff09858b8d44d9dd32871.png)
with equality on the upper bound if and only if and 
. is an integer, the claim implies that we require 
. Now, consider the minimal even 
such that 
. As a result of the equality condition noted above, we require 
and 
(since both of these are indeed valid terms). But these are consecutive terms, so one of them must have an even index. But since![\[\frac{n-2}{2} < \frac{n-2}{2}+1 < n\]](http://latex.artofproblemsolving.com/3/a/9/3a9ca2ae73e4f103b2f0fad5fcb3a7acc8fc3364.png)
for all , this implies that there exists a smaller even indexed term 
which is a contradiction. Thus, for all even , is not an integer., 
if and only if 
as a result of the equality condition. We again resort to induction. Say the only integers 
for some 
such that are those of the form for 
. Then, for 
, if and only if . But,![\[2^{k-1}-1=\frac{2^k -2}{2}<\frac{n-1}{2} \le \frac{2^{k+1}-2}{2}=2^k -1\]](http://latex.artofproblemsolving.com/e/0/a/e0a91b92240920dbd2845789c8b114a526308c3f.png)
and the only such term in this range is 
by assumption. Thus, the only such term in the range is 
which completes the induction.
for some positive integer 
.
, , andfor all integers . Determine all positive integers such thatis an integer. for all , then only possible integer solution for is or . is obvious , due to the solution present by others, also values of new sequence you give will be smaller than or equal to the original sequence.
for some bigger k and the k is not that much big and 
so the only thing that I have to care is that 
and yeah that's the place where I stuck how to prove if that is true or false. If it is false then how can I get to the contradiction
- 1
.
,
or 
.
to 
and 
otherwise.
for all 
and for the other interval using IVT we can show that 
,
.
,
and thus 
which is indeed true!
,
.
.
.
,
.
of the form 
and 
.
.
.
then, 
.
then, 
.
then, 
.
then, 
.
.
.
.
.
.
.
,
.
.
:
.
.
for all 
for all 
and 
is an integer in that given range in each case.
for all 
Implying, 
Thus, 
implying 
Thus by 
we have 
thus working 
for every positive integer 
for all 
Thus, 
, so, 
but 
by 
.
for all positive integer 
I checked cases upto 
.
.
, then the pattern is obvious and now induction. The only hurdle is 
mins of finding the values of the sequence till 
for all positive integers 
where 
.
by our inductive hypothesis
\hline
for some positive integer 
i
.
,
is even and other is odd, so by our inductive hypothesis,
is odd and the other is even.
. Hence our induction is complete.
is good.
.
must be good, similarly 
must also be good and so on.......
for all 
.
.
.
.
,
as desired, so our claim is proved.
work.
for all positive integers 
,
.![$[1,i]$](http://latex.artofproblemsolving.com/8/4/a/84a74ff446b0b9f0557add86de63bca29afb4f48.png)
.
,
by our hypothesis.
.
![$a_n \in (n, 2n]$](http://latex.artofproblemsolving.com/5/c/e/5ce1ea1ba1a13d3d5be44a90144a24dc75ed8b1b.png)
.
can't both be one less than powers of two. In the second equation, 
has equality, which gives the conclusion.
is odd.
be odd.
.
is odd.
.
.
.
.
and 
,