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such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist.
, a line 
to tangent , and another circle 
disjoint from such that and lie on opposite sides of . The tangents to from a variable point 
on meet at 
and 
. Prove that, as varies over , the circumcircle of 
is tangent to two fixed circles.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
+1 w
, 
, 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
: 
. Say it's true for 
, 
.
, 
.
: 
. Say it's true for 
.
.
.
: If then must be of the form 
, 
(Since 
)
.
, we take cases.: 
, 
.: 
, 
.
: 
. Say 
.
.
- 1
.
,
or 
.
to 
and 
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 
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 
,